Exploring Algebraic Identities shows why some algebraic rules hold for every value of their variables and how to use them. It builds (a + b)², (a − b)² and (a + b + c)² from area models, uses a² − b² for quick products, factorises quadratics with algebra tiles and by splitting the middle term, derives the cubic identities from a cube model, and simplifies rational expressions by cancelling common factors. These identities speed up calculation, turn hard factorisation into pattern matching, and are used throughout Class 10 algebra.
Key Concepts
1. Identities and Equations: What Makes a Rule True for All Values
The chapter opens with a number pattern. Take any three consecutive square numbers, for example 1, 4 and 9. Add the smallest and the largest: 1 + 9 = 10. Subtract twice the middle square: 10 − (2 × 4) = 2. Try 25, 36 and 49: (25 + 49) − (2 × 36) = 74 − 72 = 2. The answer keeps coming out as 2.
Checking a few cases makes a pattern believable. It does not prove it, because there are infinitely many sets of consecutive squares and nobody can check them all. Algebra settles the question. Let the three consecutive numbers be (n − 1), n and (n + 1). Their squares are (n − 1)2, n2 and (n + 1)2. Then
(n − 1)2 + (n + 1)2 = n2 − 2n + 1 + n2 + 2n + 1 = 2n2 + 2
Subtracting twice the middle square, 2n2, leaves 2 for every value of n. That one line is a proof. The tool that made it work is an algebraic identity.
Equation or identity. An algebraic identity is an equation that is true for all values of the variables occurring in it. An ordinary equation need not be true for all values.
| Statement | When is it true? | Type |
|---|---|---|
| x2 − 1 = 24 | Only for x = 5 or x = −5 | Equation |
| (x + y)2 = x2 + 2xy + y2 | For every value of x and y | Identity |
| (x + y)2 = x2 + y2 | Only when xy = 0 | Equation (a common false “identity”) |
How to test a claimed identity. One value that fails is enough to show a statement is not an identity. Putting in several values that work only builds confidence. To prove an identity, expand one side using the distributive property until it becomes the other side.
2. The Square of a Binomial: (a + b)2 and (a − b)2
The area model for (a + b)2. Join a line segment of length a units to one of length b units to make a segment of length (a + b) units. Build a square of side (a + b) on it. Draw one line across and one line down at the point where a ends. The big square is now cut into four pieces:
- a square of side a, area a2
- a square of side b, area b2
- two rectangles, each a by b, area ab each
The four pieces together make the whole square, so (a + b)2 = a2 + 2ab + b2.
Beyond lengths. The area picture only works when a and b are lengths, so they must be positive. The textbook checks negative and rational values too. With a = −2 and b = −3, both sides equal 25. With a = −2/3 and b = 3/4, both sides equal 1/144. The proof for all numbers uses the distributive property:
(a + b)2 = (a + b)(a + b) = a(a + b) + b(a + b) = a2 + ab + ba + b2 = a2 + 2ab + b2
The minus version. Replace b by −b in the identity above. Then 2a(−b) = −2ab and (−b)2 = b2, which gives (a − b)2 = a2 − 2ab + b2. There is also an area model. Take a square of side a and mark a length b along one side, so the rest is (a − b). The big square (area a2) now contains a small square of area (a − b)2, a larger rectangle of area ab and a smaller rectangle of area b(a − b). So
(a − b)2 = a2 − ab − b(a − b) = a2 − ab − ab + b2 = a2 − 2ab + b2
Expanding with the identity. Whatever sits in the first place is a and whatever sits in the second place is b, including its coefficient. For (5x + 2y)2, a = 5x and b = 2y:
(5x + 2y)2 = (5x)2 + 2(5x)(2y) + (2y)2 = 25x2 + 20xy + 4y2
Squaring numbers mentally. Split the number into a round number plus or minus a small one.
| Number | Split | Working | Answer |
|---|---|---|---|
| 432 | (40 + 3)2 | 1600 + 240 + 9 | 1849 |
| 292 | (30 − 1)2 | 900 − 60 + 1 | 841 |
| 2992 | (300 − 1)2 | 90000 − 600 + 1 | 89401 |
Comparing (a + b)2 with a2 + b2. With a = 10 and b = 2, (a + b)2 = 144 and a2 + b2 = 104. The expansion shows the difference is exactly 2ab, so the sign of ab decides:
- If a and b have the same sign (ab > 0), then (a + b)2 > a2 + b2.
- If a and b have opposite signs (ab < 0), then (a + b)2 < a2 + b2.
- If a = 0 or b = 0 (ab = 0), the two are equal.
3. Factorising Perfect Square Expressions
An identity can be read in both directions. Left to right it expands. Right to left it factorises: an expression of the form a2 + 2ab + b2 equals (a + b)2, and one of the form a2 − 2ab + b2 equals (a − b)2. Such expressions are called perfect squares.
A three-step check.
- Take out any common factor from all the terms first.
- Check that the first and last terms are squares, and find a and b.
- Check that the middle term is exactly 2ab. Its sign tells you whether to write (a + b)2 or (a − b)2.
| Expression | Written as a2 ± 2ab + b2 | Factors |
|---|---|---|
| x2 + 4x + 4 | (x)2 + 2(x)(2) + 22 | (x + 2)2 |
| 36x2 + 12x + 1 | (6x)2 + 2(6x)(1) + 12 | (6x + 1)2 |
| 9x2 + 24xy + 16y2 | (3x)2 + 2(3x)(4y) + (4y)2 | (3x + 4y)2 |
| 16y2 − 24y + 9 | (4y)2 − 2(4y)(3) + 32 | (4y − 3)2 |
Common factor first. In 50p2 + 60pq + 18q2 the first term 50p2 is not the square of a neat term (its square root would need √50). The textbook avoids the square root by noticing that 2 divides every term:
50p2 + 60pq + 18q2 = 2(25p2 + 30pq + 9q2) = 2[(5p)2 + 2(5p)(3q) + (3q)2] = 2(5p + 3q)2
The common factor does not have to be a whole number. For 3a2 + 4ab + (4/3)b2, take out 3: 3[a2 + (4/3)ab + (4/9)b2] = 3(a + (2/3)b)2. Here a and (2/3)b are the two terms, and 2 × a × (2/3)b = (4/3)ab, which matches.
When you factor x2 + 4x + 4 as (x + 2)2, you can also say that (x + 2) is a factor of x2 + 4x + 4.
4. The Square of a Trinomial and the Difference of Squares
Three terms. To find (a + b + c)2, treat b + c as one quantity d. Then (a + d)2 = a2 + 2ad + d2 = a2 + 2a(b + c) + (b + c)2. Expanding both brackets gives
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Read it as: every term squared once, plus twice every pair. Three terms make three pairs (ab, bc, ca).
Area model. A square of side (a + b + c), with each side divided into a, b and c, splits into three squares (a2, b2, c2) and six rectangles (two each of ab, bc and ca).
Signs travel with the terms. In (3x − 2y + 4z)2, take a = 3x, b = −2y and c = 4z:
(3x − 2y + 4z)2 = 9x2 + 4y2 + 16z2 + 2(3x)(−2y) + 2(−2y)(4z) + 2(4z)(3x) = 9x2 + 4y2 + 16z2 − 12xy − 16yz + 24zx
The squared terms are always positive. A product term is negative when exactly one of its two terms is negative.
Squaring numbers with three parts. 1192 = (100 + 10 + 9)2 = 10000 + 100 + 81 + 2000 + 1800 + 180 = 14161.
Factorising a trinomial square. Six terms, three of them perfect squares, is the signal. For 9a2 + 4b2 + c2 − 12ab + 6ac − 4bc, the squares give 3a, 2b and c. The signs of the product terms fix which one is negative: ab is negative, bc is negative and ac is positive, so b carries the minus sign. The answer is (3a − 2b + c)2.
Difference of two squares. From Grade 8, a2 − b2 = (a + b)(a − b). Rearranged, it becomes a2 = (a + b)(a − b) + b2. The chapter explains this form with a figure and credits it to Śhrīdharāchārya, who proposed it in 750 CE as a quick way to square numbers. Pick b so that a + b and a − b are easy to multiply:
| Square | (a + b)(a − b) + b2 | Answer |
|---|---|---|
| 552 | 60 × 50 + 25 | 3025 |
| 852 | 90 × 80 + 25 | 7225 |
| 1052 | 110 × 100 + 25 | 11025 |
The pattern for numbers ending in 5: multiply the number formed by the leading digits by the next number, then write 25 after it. For 85, 8 × 9 = 72, so 852 = 7225.
5. Algebra Tiles and Factorising Quadratics by Splitting the Middle Term
Algebra tiles. Three kinds of tiles stand for the terms of a quadratic expression:
- an x2-tile: a square of side x, area x2
- an x-tile: a strip of length x and width 1, area x
- a unit tile: a square of side 1, area 1
A rectangle of sides x + 3 and x + 4 has area (x + 3)(x + 4) = x2 + 3x + 4x + 12 = x2 + 7x + 12. With tiles: one x2-tile sits in a corner, three x-tiles are placed on its right, four x-tiles are placed below it, and the 12 unit tiles fill the remaining corner as a 3 by 4 array. The 7x has been split as 3x + 4x. Reading the finished rectangle gives the product (x + 3)(x + 4) = x2 + 7x + 12. Starting from the tiles for x2 + 7x + 12 and arranging them into a rectangle gives the factors.
The general pattern. (x + 3)(x + 4) = x2 + 7x + 12 and (x + 6)(x + 7) = x2 + 13x + 42. In general
(x + a)(x + b) = x2 + (a + b)x + ab
For two factors with any leading coefficients, the distributive property gives (px + a)(qx + b) = pqx2 + (pb + qa)x + ab. For example (2x + 3)(3x + 1) = 6x2 + 2x + 9x + 3 = 6x2 + 11x + 3.
Factorising without tiles. Compare the given quadratic with x2 + (a + b)x + ab. You need two numbers whose sum is the coefficient of x and whose product is the constant term.
| Expression | Sum needed | Product needed | Numbers | Factors |
|---|---|---|---|---|
| x2 + 7x + 12 | 7 | 12 | 3, 4 | (x + 3)(x + 4) |
| x2 + 11x + 30 | 11 | 30 | 5, 6 | (x + 5)(x + 6) |
| x2 − 5x + 6 | −5 | 6 | −2, −3 | (x − 2)(x − 3) |
| x2 + x − 20 | 1 | −20 | 5, −4 | (x + 5)(x − 4) |
| r2 − r − 42 | −1 | −42 | −7, 6 | (r − 7)(r + 6) |
For x2 + 11x + 30, the pairs 2 and 15 or 3 and 10 give product 30 but the wrong sum, so list the factor pairs of the constant and test each sum. Sign rules help: a positive constant means both numbers share the sign of the middle coefficient; a negative constant means the numbers have opposite signs, and the larger one (in size) takes the sign of the middle coefficient.
When the x2 coefficient is not 1. For px2 + qx + r, find two numbers whose sum is q and whose product is pr, split the middle term, and group. For 6x2 + 7x + 2: product 6 × 2 = 12, sum 7, numbers 3 and 4. So 6x2 + 3x + 4x + 2 = 3x(2x + 1) + 2(2x + 1) = (3x + 2)(2x + 1).
6. Cubic Identities
Cube of a sum. Multiply (a + b) by (a + b)2:
(a + b)3 = (a + b)(a2 + 2ab + b2) = a3 + 2a2b + ab2 + a2b + 2ab2 + b3 = a3 + 3a2b + 3ab2 + b3
The cube model. A cube of edge (a + b) has volume (a + b)3. Cut it at the point where a ends along each of its three edges. It splits into two cubes and six cuboids: one cube of volume a3, one cube of volume b3, three cuboids of size a × a × b (volume a2b each) and three cuboids of size a × b × b (volume ab2 each). The six cuboids together give 3a2b + 3ab2.
Cube of a difference. Replace b by −b: [a + (−b)]3 = a3 + 3a2(−b) + 3a(−b)2 + (−b)3, so (a − b)3 = a3 − 3a2b + 3ab2 − b3. Two terms are positive and two are negative, and the signs alternate.
Reading a cube backwards. A cube has volume p3 + 6p2q + 12pq2 + 8q3. Write it as (p)3 + 3(p)2(2q) + 3(p)(2q)2 + (2q)3. This matches a3 + 3a2b + 3ab2 + b3 with a = p and b = 2q, so the volume is (p + 2q)3 and a side of the cube is p + 2q units. In the same way 8n3 − 60n2m + 150nm2 − 125m3 = (2n)3 − 3(2n)2(5m) + 3(2n)(5m)2 − (5m)3 = (2n − 5m)3.
Sum and difference of cubes. Multiply out (x − y)(x2 + xy + y2): x3 + x2y + xy2 − x2y − xy2 − y3 = x3 − y3. The middle terms cancel in pairs. So
x3 − y3 = (x − y)(x2 + xy + y2) and x3 + y3 = (x + y)(x2 − xy + y2)
Memory aid: the sign in the linear factor matches the sign between the cubes, and the middle sign of the quadratic factor is the opposite one.
x − y as a factor. x − y divides x2 − y2 and x3 − y3. It also divides x4 − y4, because x4 − y4 = (x2 − y2)(x2 + y2) = (x − y)(x + y)(x2 + y2).
The three-cube identity. Multiplying (x + y + z) by (x2 + y2 + z2 − xy − xz − yz) term by term, all the mixed terms cancel and three copies of −xyz remain:
x3 + y3 + z3 − 3xyz = (x + y + z)(x2 + y2 + z2 − xy − yz − zx)
Two consequences follow at once:
- If x + y + z = 0, then x3 + y3 + z3 = 3xyz.
- Since x2 + y2 + z2 − xy − yz − zx = (x + y + z)2 − 3(xy + yz + zx), the identity can be written x3 + y3 + z3 − 3xyz = (x + y + z)[(x + y + z)2 − 3(xy + yz + zx)]. This is handy when you are given the sum and the sum of products.
7. Simplifying Rational Expressions and Using Factorisation in Problems
A rational expression is one algebraic expression divided by another, such as (x2 − 7x + 12)/(5x2 + 5x − 100). It is simplified the same way a fraction like 12/18 is: factorise the top and the bottom, then cancel the common factors.
Why the “not equal to zero” condition matters. You can cancel a factor only if it is not zero, because dividing by zero is not defined. The textbook states that the denominator is not equal to 0. That guarantees every factor of the denominator is non-zero, so cancelling is allowed.
Textbook Example 16. Simplify (x2 − 7x + 12)/(5x2 + 5x − 100), assuming 5x2 + 5x − 100 ≠ 0.
- Numerator: two numbers with sum −7 and product 12 are −3 and −4, so x2 − 7x + 12 = (x − 3)(x − 4).
- Denominator: take out 5 to get 5(x2 + x − 20). Two numbers with sum 1 and product −20 are 5 and −4, so the denominator is 5(x − 4)(x + 5).
- Cancel the common factor (x − 4): the result is (x − 3)/[5(x + 5)].
Factorising in word problems. Many problems give an area or a volume as an expression and ask for the sides.
- Tiles to rectangle (Saira, Example 17). A square of side x, 8 strips of x by 1 and 15 unit squares have area x2 + 8x + 15 = (x + 5)(x + 3), so the rectangle is (x + 5) by (x + 3) units.
- The pool (Example 18). A rectangular pool has breadth 4 metres less than its length and area 96 sq. metres. With length x, x(x − 4) = 96, so x2 − 4x − 96 = 0. Since (−12) × 8 = −96 and (−12) + 8 = −4, x2 − 12x + 8x − 96 = (x − 12)(x + 8) = 0. So x = 12 or x = −8. A length cannot be negative, so the length is 12 m and the breadth is 8 m.
The last step uses a key fact: if a product of two numbers is 0, at least one of them is 0. Always check the answers against the situation and reject values that make no sense, such as a negative length.
Formula and Theorem Sheet
| Identity | Statement |
|---|---|
| Square of a sum | (a + b)2 = a2 + 2ab + b2 |
| Square of a difference | (a − b)2 = a2 − 2ab + b2 |
| Square of three terms | (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca |
| Difference of squares | a2 − b2 = (a + b)(a − b) |
| Śhrīdharāchārya’s form | a2 = (a + b)(a − b) + b2 |
| Product of two binomials | (x + a)(x + b) = x2 + (a + b)x + ab |
| General linear product | (ax + b)(cx + d) = acx2 + (ad + bc)x + bd |
| Cube of a sum | (x + y)3 = x3 + 3x2y + 3xy2 + y3 |
| Cube of a difference | (x − y)3 = x3 − 3x2y + 3xy2 − y3 |
| Difference of cubes | x3 − y3 = (x − y)(x2 + xy + y2) |
| Sum of cubes | x3 + y3 = (x + y)(x2 − xy + y2) |
| Three-cube identity | x3 + y3 + z3 − 3xyz = (x + y + z)(x2 + y2 + z2 − xy − yz − zx) |
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Important Definitions
- Algebraic identity: an equation that is true for all values of the variables occurring in it, for example (x + y)2 = x2 + 2xy + y2.
- Equation: a statement of equality between two expressions that need not be true for all values of the variable. x2 − 1 = 24 is true only for x = 5 or −5.
- Factorisation: writing an algebraic expression as a product of simpler expressions, called its factors. Example: x2 + 7x + 12 = (x + 3)(x + 4).
- Perfect square expression: an expression that equals the square of another expression, such as 36x2 + 12x + 1 = (6x + 1)2.
- Quadratic expression: an expression whose highest power of the variable is 2, such as x2 − 5x + 6.
- Splitting the middle term: writing the x-term of a quadratic as a sum of two terms whose coefficients have the required sum and product, so that the expression can be grouped into factors.
- Rational expression: an expression of the form P/Q where P and Q are algebraic expressions and Q is not equal to 0.
Solved Examples (NCERT-Based)
Example 1: Expanding with fractions (Exercise Set 4.1)
Question. Using the identity (a + b)2 = a2 + 2ab + b2, expand ((7/5)x + (3/2)y)2.
Solution. Here a = (7/5)x and b = (3/2)y.
a2 = (49/25)x2
2ab = 2 × (7/5)x × (3/2)y = (42/10)xy = (21/5)xy
b2 = (9/4)y2
∴ ((7/5)x + (3/2)y)2 = (49/25)x2 + (21/5)xy + (9/4)y2
Example 2: Factoring perfect squares with fractions (Exercise Set 4.2)
Question. Factor completely: (i) 64p2 + (32/3)pq + (4/9)q2 (ii) (9/5)s2 + 6sv + 5v2
Solution. (i) 64p2 = (8p)2 and (4/9)q2 = ((2/3)q)2. Check the middle term: 2 × 8p × (2/3)q = (32/3)pq. It matches, so
64p2 + (32/3)pq + (4/9)q2 = (8p + (2/3)q)2
(ii) The textbook hint says to take out a common factor, as in Example 7. Take out 1/5: (9/5)s2 + 6sv + 5v2 = (1/5)(9s2 + 30sv + 25v2). Now 9s2 = (3s)2, 25v2 = (5v)2 and 2 × 3s × 5v = 30sv. So
(9/5)s2 + 6sv + 5v2 = (1/5)(3s + 5v)2
Example 3: Square of a three-part number (Exercise Set 4.3)
Question. Find the following square using one of the above identities: 2142.
Solution. Write 214 = 200 + 10 + 4 and use (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca.
2142 = 2002 + 102 + 42 + 2(200)(10) + 2(10)(4) + 2(4)(200)
= 40000 + 100 + 16 + 4000 + 80 + 1600 = 45796
Example 4: Is it an identity? (Exercise Set 4.3)
Question. Is this an identity? (a + b − c)2 + (a − b + c)2 + (a − b − c)2 = 2a2 + 2b2 + 2c2.
Solution. Expand each square with the three-term identity.
(a + b − c)2 = a2 + b2 + c2 + 2ab − 2bc − 2ca
(a − b + c)2 = a2 + b2 + c2 − 2ab − 2bc + 2ca
(a − b − c)2 = a2 + b2 + c2 − 2ab + 2bc − 2ca
Adding: left side = 3a2 + 3b2 + 3c2 − 2ab − 2bc − 2ca.
This is not the same as 2a2 + 2b2 + 2c2. A single check confirms it: with a = 1, b = 0, c = 0 the left side is 1 + 1 + 1 = 3 but the right side is 2. So it is not an identity.
Example 5: Completing factorisations (Exercise Set 4.4)
Question. Fill in the blanks to complete the following identities: (i) (________) (x + 1) = (3x2 − 4x − 7) (ii) 10x2 − 11x − 6 = (2x − ___) (___ + 2)
Solution. (i) The missing factor must start with 3x (to give 3x2) and end with −7 (since −7 × 1 = −7). Check: (3x − 7)(x + 1) = 3x2 + 3x − 7x − 7 = 3x2 − 4x − 7. The blank is 3x − 7.
(ii) The second bracket must start with 5x (since 2x × 5x = 10x2). The first blank must be 3 (since −3 × 2 = −6). Check: (2x − 3)(5x + 2) = 10x2 + 4x − 15x − 6 = 10x2 − 11x − 6. So 10x2 − 11x − 6 = (2x − 3)(5x + 2).
Example 6: Products without direct multiplication (Exercise Set 4.4)
Question. Select and use the identity that will help you to find the following products without multiplying directly: (23 × 17) and (34 × 43).
Solution. 23 and 17 are both 3 away from 20, so use the difference of squares:
23 × 17 = (20 + 3)(20 − 3) = 400 − 9 = 391
34 and 43 are not symmetric about a round number, so use (x + a)(x + b) = x2 + (a + b)x + ab with x = 40, a = −6, b = 3:
34 × 43 = (40 − 6)(40 + 3) = 1600 + (−6 + 3)(40) + (−6)(3) = 1600 − 120 − 18 = 1462
Example 7: Cubes of numbers (End-of-Chapter Exercises)
Question. Find the values using suitable identities: 1993 and (−107)3.
Solution. 1993 = (200 − 1)3 = 2003 − 3(200)2(1) + 3(200)(1)2 − 13 = 8000000 − 120000 + 600 − 1 = 7880599
(−107)3 = −(1073), because an odd power of a negative number is negative.
1073 = (100 + 7)3 = 1000000 + 3(10000)(7) + 3(100)(49) + 343 = 1000000 + 210000 + 14700 + 343 = 1225043
∴ (−107)3 = −1225043
Example 8: Factoring cubes (End-of-Chapter Exercises)
Question. Factor the following algebraic expressions: (i) 27b3 − 1/(64b3) (ii) 27u3 − 1/125 − (27u2)/5 + (9u)/25
Solution. (i) 27b3 = (3b)3 and 1/(64b3) = (1/(4b))3. Use x3 − y3 = (x − y)(x2 + xy + y2) with x = 3b and y = 1/(4b):
x2 = 9b2, xy = 3b × 1/(4b) = 3/4, y2 = 1/(16b2)
∴ 27b3 − 1/(64b3) = (3b − 1/(4b))(9b2 + 3/4 + 1/(16b2))
(ii) Rearrange in descending powers: 27u3 − (27u2)/5 + (9u)/25 − 1/125. The first and last terms are (3u)3 and (1/5)3. Test (a − b)3 with a = 3u and b = 1/5:
3a2b = 3 × 9u2 × (1/5) = (27u2)/5 and 3ab2 = 3 × 3u × (1/25) = (9u)/25. Both match, with signs −, + as in a3 − 3a2b + 3ab2 − b3.
∴ the expression = (3u − 1/5)3
Example 9: Simplifying a rational expression (Exercise Set 4.5)
Question. Simplify the following rational expression assuming that the expression in the denominator is not equal to zero: [(x2 + x − 6)(x2 − 7x + 12)] / [(x2 − 6x + 8)(x2 − 9)].
Solution. Factor each quadratic.
- x2 + x − 6: sum 1, product −6, numbers 3 and −2, so (x + 3)(x − 2).
- x2 − 7x + 12: sum −7, product 12, numbers −3 and −4, so (x − 3)(x − 4).
- x2 − 6x + 8: sum −6, product 8, numbers −2 and −4, so (x − 2)(x − 4).
- x2 − 9 = (x + 3)(x − 3), difference of squares.
The expression = [(x + 3)(x − 2)(x − 3)(x − 4)] / [(x − 2)(x − 4)(x + 3)(x − 3)].
Every factor on top also appears below, and none is zero because the denominator is not zero. After cancelling, the expression simplifies to 1.
Example 10: Using the three-cube identity (End-of-Chapter Exercises)
Question. If a + b + c = 5 and ab + bc + ca = 10, then prove that a3 + b3 + c3 − 3abc = −25.
Solution. First find a2 + b2 + c2:
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
25 = a2 + b2 + c2 + 20, so a2 + b2 + c2 = 5.
Now a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca) = 5 × (5 − 10) = 5 × (−5) = −25. Hence proved.
Competency-Based Questions (with answers)
1. Case-based: The tile rectangle
Meera has one square tile of side x cm, 9 rectangular strips of length x cm and width 1 cm, and 20 small squares of side 1 cm. She wants to arrange all of them, with no gaps or overlaps, into one rectangle.
(a) Write the total area as an expression. (b) Find the length and breadth of the rectangle.
Answer. (a) Area = x2 + 9x + 20 sq. cm. (b) Two numbers with sum 9 and product 20 are 4 and 5, so x2 + 9x + 20 = (x + 4)(x + 5). The rectangle is (x + 5) cm by (x + 4) cm.
2. Case-based: The walking path
A school garden is a square of side 30 m. A path of uniform width w metres is built all around the outside of the garden.
(a) Write the side of the outer square. (b) Find an expression for the area of the path. (c) Find the area of the path when w = 2.
Answer. (a) The path adds w on each side, so the outer side is (30 + 2w) m. (b) Area of path = (30 + 2w)2 − 302 = 900 + 120w + 4w2 − 900 = 4w2 + 120w sq. m. (c) With w = 2: 4(4) + 240 = 256 sq. m.
3. Source-based: Śhrīdharāchārya’s squaring method
The textbook notes that in 750 CE Śhrīdharāchārya proposed the identity a2 = (a + b)(a − b) + b2 as a method to quickly compute squares of numbers, for example 552 = (55 + 5)(55 − 5) + 52 = 60 × 50 + 25 = 3025.
(a) Show that this form follows from a2 − b2 = (a + b)(a − b). (b) Use it to find 452 and 952.
Answer. (a) Add b2 to both sides of a2 − b2 = (a + b)(a − b). (b) 452 = 50 × 40 + 25 = 2025. 952 = 100 × 90 + 25 = 9025.
4. Assertion-Reason
Assertion (A): If a = −3 and b = −4, then (a + b)2 > a2 + b2.
Reason (R): (a + b)2 − (a2 + b2) = 2ab, and ab > 0 when a and b have the same sign.
(a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
Answer: (a). (a + b)2 = 49 and a2 + b2 = 25, so A is true. R is true, and it explains A: ab = 12 > 0, so the difference 2ab = 24 is positive.
5. Assertion-Reason
Assertion (A): x2 − 1 = 24 is an algebraic identity.
Reason (R): An algebraic identity is an equation that is true for all values of the variables occurring in it.
Options (a) to (d) as in Question 4.
Answer: (d). A is false: x2 − 1 = 24 holds only for x = 5 or x = −5 (for x = 0 the left side is −1). R is the correct definition, so R is true.
6. Assertion-Reason
Assertion (A): x − y is a factor of x4 − y4.
Reason (R): x4 − y4 = (x2 − y2)(x2 + y2).
Options (a) to (d) as in Question 4.
Answer: (a). Both are true. Since x2 − y2 = (x − y)(x + y), R gives x4 − y4 = (x − y)(x + y)(x2 + y2), which shows x − y is a factor. So R explains A.
7. Reasoning: Who is correct?
The textbook describes James and Reshma expanding (a − b)2 (a + b). James writes (a2 − 2ab + b2)(a + b). Reshma writes (a − b)[(a − b)(a + b)] = (a − b)(a2 − b2). According to you, who is correct and why?
Answer. Both are correct. Multiplication can be grouped in any order, so both routes give the same product.
James: (a2 − 2ab + b2)(a + b) = a3 + a2b − 2a2b − 2ab2 + ab2 + b3 = a3 − a2b − ab2 + b3.
Reshma: (a − b)(a2 − b2) = a3 − ab2 − a2b + b3 = a3 − a2b − ab2 + b3.
8. Case-based: The storage cube
A storage cube has volume p3 + 6p2q + 12pq2 + 8q3 cubic units.
(a) Find the edge of the cube. (b) Find the edge and the volume when p = 2 and q = 1, and check the volume by substituting in the given expression.
Answer. (a) p3 + 3(p)2(2q) + 3(p)(2q)2 + (2q)3 = (p + 2q)3, so the edge is p + 2q units. (b) Edge = 2 + 2 = 4 units, volume = 64 cubic units. Check: 8 + 24 + 24 + 8 = 64.
Important Questions for Exams
1-Mark Questions
Q1. Write the expansion of (3x − 2)2.
Answer. 9x2 − 12x + 4.
Q2. Find the value of 102 × 98 using an identity.
Answer. (100 + 2)(100 − 2) = 10000 − 4 = 9996.
Q3. If x + y + z = 0, what is x3 + y3 + z3 equal to?
Answer. 3xyz.
Q4. Is (x + y)2 = x2 + y2 an identity? Give a reason.
Answer. No. For x = 1, y = 1 the left side is 4 and the right side is 2.
2-Mark Questions
Q5. Factor 49g2 + 14gh + h2.
Answer. 49g2 = (7g)2, h2 = (h)2 and 2(7g)(h) = 14gh. So 49g2 + 14gh + h2 = (7g + h)2.
Q6. Find 1172 using a suitable identity.
Answer. (100 + 10 + 7)2 = 10000 + 100 + 49 + 2000 + 140 + 1400 = 13689.
Q7. Factor s2 − 11s + 24.
Answer. Two numbers with sum −11 and product 24 are −3 and −8. So s2 − 11s + 24 = (s − 3)(s − 8).
Q8. If a number plus its reciprocal equals 10/3, find the number.
Answer. x + 1/x = 10/3 gives 3x2 + 3 = 10x, so 3x2 − 10x + 3 = 0. Split −10x as −9x − x: 3x(x − 3) − 1(x − 3) = (3x − 1)(x − 3) = 0. So x = 3 or x = 1/3. Both work, since 3 + 1/3 = 10/3.
3-Mark Questions
Q9. Expand (3x − 2y + 4z)2.
Answer. With a = 3x, b = −2y, c = 4z: 9x2 + 4y2 + 16z2 + 2(3x)(−2y) + 2(−2y)(4z) + 2(4z)(3x) = 9x2 + 4y2 + 16z2 − 12xy − 16yz + 24zx.
Q10. Simplify (s3 + 125t3)/(s2 − 2st − 35t2), assuming the denominator is not 0.
Answer. Numerator: s3 + (5t)3 = (s + 5t)(s2 − 5st + 25t2). Denominator: numbers with sum −2 and product −35 are −7 and 5, so (s − 7t)(s + 5t). Cancel (s + 5t): the answer is (s2 − 5st + 25t2)/(s − 7t).
Q11. Find the value of x3 + y3 − 12xy + 64, when x + y = −4.
Answer. Write 64 = 43 and 12xy = 3 × x × y × 4. The expression is x3 + y3 + 43 − 3(x)(y)(4). Here x + y + 4 = −4 + 4 = 0, so by the three-cube identity the expression equals (x + y + 4)(…) = 0.
5-Mark Questions
Q12. Three numbers have sum 10, product 25 and the sum of their squares is 38. Find the sum of their cubes.
Answer. Let the numbers be x, y, z. (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx) gives 100 = 38 + 2(xy + yz + zx), so xy + yz + zx = 31. By the three-cube identity, x3 + y3 + z3 − 3xyz = (x + y + z)(x2 + y2 + z2 − xy − yz − zx) = 10(38 − 31) = 70. So x3 + y3 + z3 = 70 + 3(25) = 145.
Q13. If both x − 2 and x − 1/2 are factors of px2 + 5x + r, show that p = r.
Answer. If x − 2 is a factor, px2 + 5x + r = (x − 2) × (another expression), so the value is 0 when x = 2: 4p + 10 + r = 0 … (1). In the same way, x = 1/2 gives p/4 + 5/2 + r = 0, and multiplying by 4, p + 10 + 4r = 0 … (2). From (1), r = −4p − 10. Substituting in (2): p + 10 − 16p − 40 = 0, so −15p = 30 and p = −2. Then r = 8 − 10 = −2. So p = r (= −2).
Common Mistakes and Examiner Tips
- Writing (a + b)2 = a2 + b2. This drops the middle term 2ab. Fix: say “square, twice the product, square” each time, or picture the square with its two ab rectangles.
- Wrong sign on b2 in (a − b)2. Students write a2 − 2ab − b2. Fix: (−b)2 = +b2, so only the middle term changes sign.
- Losing signs in (a + b + c)2. In (3x − 2y + 4z)2, the term 2bc is 2(−2y)(4z) = −16yz. Fix: write a, b, c with their signs before substituting.
- Forgetting the common factor. Trying to factor 50p2 + 60pq + 18q2 directly leads to square roots. Fix: always take out the common factor first, then look for the pattern.
- Factorising without checking the middle term. 9x2 + 12x + 16 is not (3x + 4)2, because 2(3x)(4) = 24x. Fix: check 2ab every time before writing the square.
- Wrong pair in splitting the middle term. For x2 + 11x + 30, 3 and 10 give the right product but the wrong sum. Fix: list factor pairs of the constant and test the sum; for a negative constant, pick opposite signs.
- Mixing up the sum and difference of cubes. x3 + y3 is (x + y)(x2 − xy + y2), not (x + y)(x2 + xy + y2). Fix: the linear factor copies the sign between the cubes; the quadratic’s middle sign is the opposite.
- Using 2xy instead of xy in the cube factors. x3 − y3 = (x − y)(x2 + xy + y2). Writing 2xy turns the second factor into a perfect square, which is wrong. Fix: expand once to see the terms cancel.
- Cancelling terms instead of factors. (x2 + 9)/(x + 3) does not simplify to x + 3. Fix: factorise completely first; cancel only whole factors that appear in both numerator and denominator.
- Ignoring the non-zero condition. Write “since the denominator is not zero, this factor is not zero” before cancelling. Fix: state this reason in your written answer.
- Keeping impossible answers in word problems. In the pool problem, x = −8 must be rejected because a length cannot be negative. Fix: after solving, test every root against the situation and state why you reject any.
Quick Revision Points
- An identity is true for all values of its variables; an equation may be true only for some.
- (a + b)2 = a2 + 2ab + b2: a square of side a + b splits into a2, b2 and two ab rectangles.
- (a − b)2 = a2 − 2ab + b2: replace b by −b in the first identity.
- Square numbers by splitting: 432 = (40 + 3)2 = 1849; 292 = (30 − 1)2 = 841.
- Perfect square check: first and last terms are squares, middle term is exactly ±2ab.
- (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca: each square once, each pair twice.
- a2 − b2 = (a + b)(a − b); Śhrīdharāchārya (750 CE): a2 = (a + b)(a − b) + b2.
- Algebra tiles: x2-tile, x-tile, unit tile; x2 + 7x + 12 forms an (x + 3) by (x + 4) rectangle.
- (x + a)(x + b) = x2 + (a + b)x + ab.
- Splitting the middle term: sum = coefficient of x, product = constant (or pr when the x2 coefficient is p).
- (a + b)3 = a3 + 3a2b + 3ab2 + b3: a cube splits into 2 cubes and 6 cuboids.
- (a − b)3 = a3 − 3a2b + 3ab2 − b3: signs alternate.
- x3 − y3 = (x − y)(x2 + xy + y2); x3 + y3 = (x + y)(x2 − xy + y2).
- x3 + y3 + z3 − 3xyz = (x + y + z)(x2 + y2 + z2 − xy − yz − zx).
- If x + y + z = 0, then x3 + y3 + z3 = 3xyz.
- Rational expressions: factorise fully, cancel common factors, and only when they are not zero.
Weightage and Exam Pattern
Class 9 has no board exam. This chapter is assessed in school exams: periodic tests, the half-yearly exam and the annual exam. It is also the foundation for Class 10 algebra, where factorising quadratics and working with identities are used again and again, for example in quadratic equations.
In the CBSE Class IX Mathematics course structure for 2026-27, the theory paper is of 3 hours and 80 marks, and internal assessment carries 20 marks (pen-paper tests and multiple assessment 10, portfolio 5, lab practical 5). Exploring Algebraic Identities sits in Unit II, Algebra, which carries 20 marks together with Introduction to Polynomials, Sequences and Progressions, and Linear Equations in Two Variables. The course structure allots 66 periods to the whole Algebra unit. It does not split the 20 marks between the four chapters.
| Unit | Marks (out of 80) |
|---|---|
| I. Number System | 07 |
| II. Algebra (includes this chapter) | 20 |
| III. Coordinate Geometry | 04 |
| IV. Geometry | 25 |
| V. Mensuration | 14 |
| VI. Statistics and Probability | 10 |
The CBSE question paper design weights questions by typology: 43 marks (about 54%) for remembering and understanding, 19 marks (about 24%) for applying, and 18 marks (about 22%) for analysing, evaluating and creating.
The course structure lists the key concepts as: revisiting identities, visualising them with geometrical models, factorisation using identities, more identities, factorising quadratics with and without algebra tiles, finding new identities, and simplifying rational expressions.
Class 9 Maths · Chapter 4 – swipe through all 10 cards to understand the whole chapter.
Identity vs Equation
An identity is true for every value of its variables; an equation may be true only for some.
One failing value disproves a claimed identity.
- x2 − 1 = 24 is true only for x = 5 or −5, so it is an equation
- Prove identities by expanding with the distributive property
- Consecutive squares: (n − 1)2 + (n + 1)2 − 2n2 = 2 always
Square of a Sum and a Difference
A square of side a + b splits into a2, b2 and two ab rectangles.
Replace b by −b to get the minus version.
- 432 = (40 + 3)2 = 1600 + 240 + 9 = 1849
- 292 = (30 − 1)2 = 900 − 60 + 1 = 841
- (a + b)2 − (a2 + b2) = 2ab, so the sign of ab decides which is bigger
Factorising Perfect Squares
Read a2 ± 2ab + b2 backwards to write it as (a ± b)2.
Take out any common factor first.
- 36x2 + 12x + 1 = (6x + 1)2
- 50p2 + 60pq + 18q2 = 2(5p + 3q)2
- Always check the middle term equals exactly 2ab
Square of a Trinomial
Every term squared once, plus twice every pair.
Signs travel with the terms.
- Proof: put d = b + c and use (a + d)2
- 1192 = (100 + 10 + 9)2 = 14161
- 9a2 + 4b2 + c2 − 12ab + 6ac − 4bc = (3a − 2b + c)2
a2 − b2 and Śhrīdharāchārya
a2 − b2 = (a + b)(a − b), rewritten as a2 = (a + b)(a − b) + b2 for quick squares.
Proposed by Śhrīdharāchārya in 750 CE.
- 552 = 60 × 50 + 25 = 3025
- 104 × 96 = 1002 − 42 = 9984
- 23 × 17 = 202 − 32 = 391
Tiles and the Middle Term
Arrange x2 tiles, x-tiles and unit tiles into a rectangle to read off the factors.
Sum = coefficient of x, product = constant term.
- x2 + 7x + 12 = (x + 3)(x + 4), with 7x split as 3x + 4x
- x2 − 5x + 6 = (x − 2)(x − 3)
- 6x2 + 7x + 2: split 7x as 3x + 4x to get (3x + 2)(2x + 1)
Cube of a Sum and a Difference
A cube of edge a + b splits into 2 cubes and 6 cuboids.
In (a − b)3 the signs alternate +, −, +, −.
- Cuboids: three a × a × b and three a × b × b
- p3 + 6p2q + 12pq2 + 8q3 = (p + 2q)3
- 1993 = (200 − 1)3 = 7880599
Sum and Difference of Cubes
Two cubes factor into a linear and a quadratic factor.
The middle term is xy, never 2xy.
- x − y is a factor of x2 − y2, x3 − y3 and x4 − y4
- 27b3 − 1/(64b3) = (3b − 1/(4b))(9b2 + 3/4 + 1/(16b2))
- s3 + 125t3 = (s + 5t)(s2 − 5st + 25t2)
The x3 + y3 + z3 − 3xyz Identity
Links the sum of cubes to the sum, the sum of squares and the product.
If x + y + z = 0, then x3 + y3 + z3 = 3xyz.
- Sum 10, product 25, squares 38 gives sum of cubes 145
- a + b + c = 5, ab + bc + ca = 10 gives a3 + b3 + c3 − 3abc = −25
- x3 + y3 − 12xy + 64 = 0 when x + y = −4
Simplifying Rational Expressions
Factorise the top and bottom, then cancel common factors that are not zero.
Cancel factors, never individual terms.
- Take out numerical common factors first
- (4x2 + 4x + 1)/(4x2 − 1) = (2x + 1)/(2x − 1)
- Pool of area 96 m2, breadth 4 m less: length 12 m, breadth 8 m
📝 Practice Exploring Algebraic Identities - 10 questions
Concept-check questions · with answers & explanations
Start →Close ✕
Why B: Step 1 ((a + b)² identity): 9² = a² + b² + 2ab, so 81 = a² + b² + 28. Step 2 (rearrange): a² + b² = 81 – 28 = 53. (Check: a = 7, b = 2 gives 49 + 4 = 53.)
Why not A: 67 = 81 – 14 subtracts ab, not 2ab.
Why not C: 95 = 81 + 14 adds instead of subtracting, and also uses ab.
Why not D: 25 = 81 – 56 is (a – b)², since it subtracts 4ab.
Remember: a² + b² = (a + b)² – 2ab.
Why C: The assertion is true: (n – 1)² + (n + 1)² = (n² – 2n + 1) + (n² + 2n + 1) = 2n² + 2, and subtracting 2n² leaves 2 (for 25, 36, 49: 74 – 72 = 2). The reason is false: (n – 1)² = n² – 2n + 1, not n² – 1 (n = 3 gives 4, not 8).
Why not A: R is false, so it cannot explain A.
Why not B: B needs both statements true, but R fails for n = 3.
Why not D: A is actually true for every n, as the algebra shows.
Remember: (n – 1)² = n² – 2n + 1, and the -2n and +2n cancel in the pattern.
Why D: The squares come from 3a, 2b and c. The ab term is negative, so 3a and 2b have opposite signs; the ac term is positive, so 3a and c have the same sign; then bc is negative, which fits. With x = 3a, y = -2b, z = c: 2xy = -12ab, 2yz = -4bc, 2zx = 6ac. So the expression is (3a – 2b + c)².
Why not A: all plus signs would give +12ab and +4bc.
Why not B: (-2b)(-c) gives +4bc, and (3a)(-c) gives -6ac, both wrong.
Why not C: (2b)(c) gives +4bc and (-3a)(c) gives -6ac, both wrong.
Remember: A negative pair term means those two parts carry opposite signs.
Why A: (2x + 1)(3x + 2) = 6x² + 4x + 3x + 2 = 6x² + 7x + 2. In splitting form, 7x = 3x + 4x because 3 × 4 = 6 × 2 = 12.
Why not B: (2x + 2)(3x + 1) = 6x² + 2x + 6x + 2 = 6x² + 8x + 2.
Why not C: (6x + 1)(x + 2) = 6x² + 13x + 2.
Why not D: (2x – 1)(3x – 2) = 6x² – 7x + 2, the x term has the wrong sign.
Remember: The middle term comes from the outer plus the inner products.
Why B: Step 1 (square of a trinomial): 10² = 38 + 2(xy + yz + zx), so xy + yz + zx = 62 / 2 = 31. Step 2 (three-cube identity): x³ + y³ + z³ – 3(25) = 10(38 – 31) = 70. Step 3 (add back 3xyz): x³ + y³ + z³ = 70 + 75 = 145.
Why not A: 455 = 380 + 75 ignores the xy + yz + zx part entirely.
Why not C: 380 = 10 × 38 uses only x² + y² + z² and forgets 3xyz as well.
Why not D: -165 comes from taking xy + yz + zx = 62, forgetting to halve.
Remember: Sum of pair products = [(sum)² – (sum of squares)] / 2.
Why B: Numerator: p⁴ – 16 = (p² – 4)(p² + 4) = (p – 2)(p + 2)(p² + 4). Denominator: p² – 4p + 4 = (p – 2)². One (p – 2) cancels, leaving (p² + 4)(p + 2) / (p – 2).
Why not A: the factors p – 2 and p + 2 are in the wrong places.
Why not C: the numerator has only one (p – 2), so only one of the two in the denominator can cancel.
Why not D: this treats p⁴ – 16 as (p² – 4)², which equals p⁴ – 8p² + 16.
Remember: p⁴ – 16 = (p² – 4)(p² + 4), and p² – 4 splits again.
Why A: (a + b)² = a² + b² + 2ab, so (a + b)² – (a² + b²) = 2ab. The square is smaller exactly when 2ab < 0, which happens when a and b have opposite signs.
Why not B: both positive gives ab > 0, so (a + b)² is larger, as in 12² = 144 > 104.
Why not C: both negative still gives ab > 0 (for example a = -2, b = -3: 25 > 13).
Why not D: with a = 3, b = -1: (a + b)² = 4 but a² + b² = 10, so it is not always greater.
Remember: The sign of 2ab decides which is bigger.
Why B: Every term is divisible by 2: 50p² + 60pq + 18q² = 2(25p² + 30pq + 9q²). Inside, 25p² = (5p)², 9q² = (3q)², 30pq = 2(5p)(3q), so the bracket is (5p + 3q)². Answer 2(5p + 3q)².
Why not A: (5p + 3q)² = 25p² + 30pq + 9q², only half the given expression.
Why not C: 25p and 9q are the squares, not the square roots.
Why not D: (10p + 6q)² = 4(5p + 3q)² = 100p² + 120pq + 36q², twice too big.
Remember: Take out the common factor first, then look for a² + 2ab + b².
Why C: Step 1 (choose b = 5): a + b = 70 and a – b = 60. Step 2 (product): 70 × 60 = 4200. Step 3 (add b²): 4200 + 25 = 4225.
Why not A: 4200 forgets to add b² at the end.
Why not B: 4205 adds b = 5 instead of b² = 25.
Why not D: 3625 uses 60 × 60 + 25, taking a – b twice.
Remember: Numbers ending in 5 square to (tens part) × (tens part + 1), then 25.
Why D: Step 1 (form the equation): length x, breadth x – 4, so x(x – 4) = 96, that is x² – 4x – 96 = 0. Step 2 (split the middle term, sum -4 and product -96): -12 and 8, so (x – 12)(x + 8) = 0. Step 3 (reject the negative root): x = 12 m, since a length cannot be -8. Breadth = 8 m, and 12 × 8 = 96.
Why not A: 8 m is the breadth, not the length.
Why not B: 10 m is close to the square root of 96, which assumes a square pool.
Why not C: 24 m is 96 divided by 4, which uses the difference as a side.
Remember: Always check the answer fits the story: 12 × 8 = 96.
Chapter Navigation
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Related Chapters in Class 9 Maths
- Coordinate Geometry Class 9 Notes
- Introduction to Polynomials Class 9 Notes
- Number System Class 9 Notes
- Circles Class 9 Notes
- Area and Perimeter Class 9 Notes
- Introduction to Probability Class 9 Notes
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Frequently Asked Questions
An algebraic identity is an equation that is true for all values of the variables in it, such as (x + y)² = x² + 2xy + y². An equation need not be true for all values: x² − 1 = 24 holds only for x = 5 or x = −5. One value that fails is enough to show a statement is not an identity.
A square of side a + b splits into a square a², a square b² and two rectangles of area ab. Leaving out the rectangles leaves out 2ab. So (a + b)² = a² + 2ab + b². The two sides of (a + b)² = a² + b² are equal only when ab = 0.
Find two numbers whose sum is the coefficient of x and whose product is the constant term. For x² + 11x + 30 the numbers are 5 and 6, so it factors as (x + 5)(x + 6). If the constant is negative the two numbers have opposite signs, as in r² − r − 42 = (r − 7)(r + 6).
Algebra tiles are an x² tile (a square of side x), an x-tile (a strip x by 1) and a unit tile (a square of side 1). Arranging the tiles of a quadratic into a rectangle shows its factors: one x² tile, 7 x-tiles and 12 unit tiles form a rectangle of sides x + 3 and x + 4.
x³ − y³ = (x − y)(x² + xy + y²) and x³ + y³ = (x + y)(x² − xy + y²). The sign in the linear factor matches the sign between the cubes, and the middle sign of the quadratic factor is the opposite one. The middle term is xy, not 2xy.
It equals (x + y + z)(x² + y² + z² − xy − yz − zx). It finds the sum of cubes when the sum, product and sum of squares of three numbers are known. It also gives a shortcut: if x + y + z = 0, then x³ + y³ + z³ = 3xyz.
Factorise the numerator and the denominator completely, then cancel the factors common to both. A factor can be cancelled only when it is not zero, which is why the question states that the denominator is not equal to 0. Terms joined by + or − can never be cancelled on their own.
Class 9 has no board exam, so the chapter is tested in school exams. It is part of Unit II, Algebra, which carries 20 of the 80 theory marks in the CBSE Class IX course structure, shared with three other chapters. Expect expansions, quick squares, factorisation, rational expressions and area or volume word problems.
