NCERT Solutions for Class 9th Maths Chapter 4 Chapter 4 Exploring Algebraic Identities — End-of-Chapter Exercises

Book page 88–90 Updated on2026-09-08

Q1.

Use suitable identities to find the following products: (i) (−3x + 4)² (ii) (2s + 7)(2s − 7) (iii) (p² + 1/2)(p² − 1/2) (iv) (2n + 7)(2n − 7) (v) (s − 2t)(s² + 2st + 4t²) (vi) (1/2r − 4r)² (vii) (−3m + 4k − l)² (viii) (x − y/3)³ (ix) (7k/2 − 2m/3)³

Answer

(i) (−3x + 4)² — read it as (4 − 3x)² and use (a − b)².

= 4² − 2(4)(3x) + (3x)² = 9x² − 24x + 16

(ii) (2s + 7)(2s − 7) — difference of two squares.

= (2s)² − 7² = 4s² − 49

(iii) (p² + 1/2)(p² − 1/2)

= (p²)² − (1/2)² = p⁴ − 1/4

(iv) (2n + 7)(2n − 7)

= (2n)² − 7² = 4n² − 49

(v) (s − 2t)(s² + 2st + 4t²) — this is (x − y)(x² + xy + y²) with x = s, y = 2t.

Check the partner bracket: x² = s², xy = s(2t) = 2st, y² = 4t² ✓
= s³ − (2t)³ = s³ − 8t³

(vi) (1/2r − 4r)² — with a = 1/(2r), b = 4r, note 2ab = 2 × (1/2r) × 4r = 4.

= (1/2r)² − 2(1/2r)(4r) + (4r)²
= 1/(4r²) − 4 + 16r²

(vii) (−3m + 4k − l)² — three-letter square with a = −3m, b = 4k, c = −l.

= 9m² + 16k² + l² + 2(−3m)(4k) + 2(4k)(−l) + 2(−l)(−3m)
= 9m² + 16k² + l² − 24km − 8kl + 6lm

(viii) (x − y/3)³ — cube of a difference, a = x, b = y/3.

= x³ − 3x²(y/3) + 3x(y/3)² − (y/3)³
= x³ − x²y + xy²/3 − y³/27

(ix) (7k/2 − 2m/3)³ — a = 7k/2, b = 2m/3.

a³ = 343k³/8
3a²b = 3 × (49k²/4) × (2m/3) = 49k²m/2
3ab² = 3 × (7k/2) × (4m²/9) = 14km²/3
b³ = 8m³/27

= 343k³/8 − 49k²m/2 + 14km²/3 − 8m³/27

Tip: Parts (ii) and (iv) are the same expression with a different letter. Spotting that saves you the second calculation — identities do not care what the letters are called.

Q2.

Find the values using suitable identities: (i) 17 × 21 (ii) 104 × 96 (iii) 24 × 16 (iv) 147³ (v) 199³ (vi) 127³ (vii) (−107)³ (viii) (−299)³

Answer

(i) 17 × 21 — both near 19, but at different distances, so use (x + a)(x + b) with x = 20.

= (20 − 3)(20 + 1) = 400 + (−3 + 1)(20) + (−3)(1)
= 400 − 40 − 3 = 357

(ii) 104 × 96 — equally spaced about 100, so use a² − b².

= (100 + 4)(100 − 4) = 10000 − 16 = 9984

(iii) 24 × 16 — equally spaced about 20.

= (20 + 4)(20 − 4) = 400 − 16 = 384

(iv) 147³ = (150 − 3)³

= 150³ − 3(150)²(3) + 3(150)(3)² − 3³
= 3375000 − 202500 + 4050 − 27
= 3176523

(v) 199³ = (200 − 1)³

= 8000000 − 3(40000)(1) + 3(200)(1) − 1
= 8000000 − 120000 + 600 − 1 = 7880599

(vi) 127³ = (130 − 3)³

= 2197000 − 3(16900)(3) + 3(130)(9) − 27
= 2197000 − 152100 + 3510 − 27 = 2048383

(vii) (−107)³ = −107³ = −(100 + 7)³

107³ = 1000000 + 3(10000)(7) + 3(100)(49) + 343
= 1000000 + 210000 + 14700 + 343 = 1225043
so (−107)³ = −1225043

(viii) (−299)³ = −299³ = −(300 − 1)³

299³ = 27000000 − 3(90000)(1) + 3(300)(1) − 1
= 27000000 − 270000 + 900 − 1 = 26730899
so (−299)³ = −26730899

Why an odd power keeps the sign: (−a)³ = (−1)³a³ = −a³. Cubing a negative number gives a negative answer, so compute the cube of the positive number and put the minus back at the end. (An even power would have absorbed the sign, as in (−3x + 4)² above.)

Q3.

Factor the following algebraic expressions: (i) 4y² + 1 + 1/(16y²) (ii) 9m² − 1/(25n²) (iii) 27b³ − 1/(64b³) (iv) x² + 5x/6 + 1/6 (v) 27u³ − 1/125 − 27u²/5 + 9u/25 (vi) 64y³ + z³/125 (vii) p³ + 27q³ + r³ − 9pqr (viii) 9m² − 12m + 4 (ix) 9x³ − (8/3)y³ + z³/3 + 6xyz (x) 4x² + 9y² + 36z² + 12xz + 36yz + 24xy (xi) 27u³ − 1/216 − 9u²/2 + u/4

Answer

(i) 4y² + 1 + 1/(16y²) — the middle term 1 is the giveaway: 2 × 2y × 1/(4y) = 1.

= (2y)² + 2(2y)(1/4y) + (1/4y)² = (2y + 1/(4y))²

(ii) 9m² − 1/(25n²) — difference of two squares.

= (3m)² − (1/5n)² = (3m + 1/(5n))(3m − 1/(5n))

(iii) 27b³ − 1/(64b³) — difference of two cubes, x = 3b, y = 1/(4b).

= (3b)³ − (1/4b)³
= (3b − 1/(4b))(9b² + 3/4 + 1/(16b²))
(the middle term is xy = 3b × 1/4b = 3/4)

(iv) x² + 5x/6 + 1/6 — need a + b = 5/6 and ab = 1/6; take a = 1/2, b = 1/3.

= (x + 1/2)(x + 1/3) — or, clearing fractions, (2x + 1)(3x + 1)/6

(v) 27u³ − 27u²/5 + 9u/25 − 1/125 — four terms with alternating signs: a cube of a difference.

a = 3u, b = 1/5:
a³ = 27u³, 3a²b = 3(9u²)(1/5) = 27u²/5, 3ab² = 3(3u)(1/25) = 9u/25, b³ = 1/125 ✓
= (3u − 1/5)³

(vi) 64y³ + z³/125 — sum of two cubes, x = 4y, y′ = z/5.

= (4y)³ + (z/5)³
= (4y + z/5)(16y² − 4yz/5 + z²/25)

(vii) p³ + 27q³ + r³ − 9pqr — the three-cubes identity, with 3q in place of the second letter.

= p³ + (3q)³ + r³ − 3(p)(3q)(r)
= (p + 3q + r)(p² + 9q² + r² − 3pq − 3qr − rp)

(viii) 9m² − 12m + 4

= (3m)² − 2(3m)(2) + 2² = (3m − 2)²

(ix) 9x³ − (8/3)y³ + z³/3 + 6xyz — the coefficients are not cubes, so take out 1/3 first.

= (1/3)(27x³ − 8y³ + z³ + 18xyz)
= (1/3)[ (3x)³ + (−2y)³ + z³ − 3(3x)(−2y)(z) ]
= (1/3)(3x − 2y + z)(9x² + 4y² + z² + 6xy + 2yz − 3zx)

(x) 4x² + 9y² + 36z² + 12xz + 36yz + 24xy — the three squares point to 2x, 3y and 6z, and their cross terms are 2(2x)(3y) = 12xy, 2(3y)(6z) = 36yz, 2(6z)(2x) = 24zx. So the expression that factors is

4x² + 9y² + 36z² + 12xy + 36yz + 24zx = (2x + 3y + 6z)²

Note on the printed question: the English edition prints this part as “12xz + 36yz + 24xy” — the coefficients 12 and 24 attached to the wrong pairs. With those coefficients the expression is not a perfect square and does not factor at all (put x = y = z = 1: it gives 4 + 9 + 36 + 12 + 36 + 24 = 121 = 11², but x = 1, y = 1, z = 0 gives 4 + 9 + 24 = 37, which is not a square of 2 + 3 = 5). The Hindi edition prints the intended form, 12xy + 36yz + 24zx, and that is the version factored above.

(xi) 27u³ − 9u²/2 + u/4 − 1/216 — again a cube of a difference.

a = 3u, b = 1/6:
3a²b = 3(9u²)(1/6) = 9u²/2 ✓ 3ab² = 3(3u)(1/36) = u/4 ✓ b³ = 1/216 ✓
= (3u − 1/6)³

How to tell a cube from a square at a glance: a perfect-square trinomial has three terms and the outer two are squares; a perfect-cube expansion has four terms, the outer two are cubes, and the two middle coefficients are in the ratio 3a²b : 3ab² = a : b. In (v), 27u²/5 : 9u/25 = 15u : 1, and indeed a : b = 3u : 1/5 = 15u : 1. ✓

Q4.

Simplify the following: (i) (4x² + 4x + 1)/(4x² − 1) (ii) 9(3a³ − 24b³)/(9a² − 36b²) (iii) (s³ + 125t³)/(s² − 2st − 35t²). Note: Assume that the denominators are not equal to 0.

Answer

(i)

Numerator = (2x)² + 2(2x)(1) + 1² = (2x + 1)²
Denominator = (2x)² − 1² = (2x + 1)(2x − 1)

= (2x + 1)² / [(2x + 1)(2x − 1)] = (2x + 1)/(2x − 1)

(ii) — take out the numerical factors first, then use the difference of cubes and the difference of squares.

Numerator = 9 × 3(a³ − 8b³) = 27(a − 2b)(a² + 2ab + 4b²)
Denominator = 9(a² − 4b²) = 9(a − 2b)(a + 2b)

= 27(a − 2b)(a² + 2ab + 4b²) / [9(a − 2b)(a + 2b)]
= 3(a² + 2ab + 4b²)/(a + 2b)

(iii)

Numerator = s³ + (5t)³ = (s + 5t)(s² − 5st + 25t²)
Denominator = s² − 2st − 35t² = (s − 7t)(s + 5t)

= (s² − 5st + 25t²)/(s − 7t)

Tip: In (ii) resist the urge to cancel the 9s straight away. Factor completely first — the useful cancellation is (a − 2b), and you would not have seen it if you had stopped at 3(a³ − 8b³)/(a² − 4b²).

Q5.

Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units. (i) 25a² − 30ab + 9b² (ii) 36s² − 49t²

Answer

Area = length × breadth, so factorising the area into two factors gives a possible pair of dimensions.

(i) 25a² − 30ab + 9b²

= (5a)² − 2(5a)(3b) + (3b)² = (5a − 3b)²
Length = (5a − 3b) units, Breadth = (5a − 3b) units

Both sides are equal, so this rectangle is in fact a square.

(ii) 36s² − 49t²

= (6s)² − (7t)² = (6s + 7t)(6s − 7t)
Length = (6s + 7t) units, Breadth = (6s − 7t) units

Why the answer says ‘possible’: a given area can be split in many ways — 2(3s + 3.5t)(6s − 7t) has the same product. The question asks for a factorisation into sensible algebraic expressions, and for that the identities give the natural one. For (ii), taking the larger factor as the length is the usual convention, since 6s + 7t > 6s − 7t whenever t > 0.

Q6.

Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units. (i) 6a² − 24b² (ii) 3ps² − 15ps + 12p

Answer

Volume = length × breadth × height, so factor each expression into three factors.

(i) 6a² − 24b²

= 6(a² − 4b²)
= 6(a − 2b)(a + 2b)
Length = 6 units, Breadth = (a + 2b) units, Height = (a − 2b) units

(ii) 3ps² − 15ps + 12p

= 3p(s² − 5s + 4)
Need two numbers with sum −5 and product 4: they are −1 and −4
= 3p(s − 1)(s − 4)
Length = 3p units, Breadth = (s − 1) units, Height = (s − 4) units

Tip: Always pull out the common numerical or letter factor first — in (i) it supplies one whole dimension, and in (ii) it supplies 3p. What is left is then a quadratic you already know how to split.

Q7.

The village playground is shaped as a square of side 40 metres. A path of width s metres is created around the playground for people to walk. Find an expression for the area of the path in terms of s.

Answer

The path is the region between two squares: the outer square (playground + path on both sides) and the playground itself.

The playground, 40 m square, with a path of width s all round. The outer square has side 40 + 2s.

Side of outer square = 40 + s + s = (40 + 2s) m
Area of outer square = (40 + 2s)² = 40² + 2(40)(2s) + (2s)²
= 1600 + 160s + 4s²
Area of playground = 40² = 1600

Area of path = (1600 + 160s + 4s²) − 1600
= (4s² + 160s) m² = 4s(s + 40) m²

Why the side gains 2s and not s: the path runs all the way round, so it adds s metres on the left and s metres on the right. The same happens top and bottom. Forgetting the second s is the commonest slip in path problems.

Check it yourself: take s = 1. The path should be a 1 m border round a 40 m square: four strips of 40 × 1 plus four corner squares of 1 × 1, that is 160 + 4 = 164 m². The formula gives 4(1) + 160(1) = 164 m². ✓ The 4s² is exactly the four corners.

Q8.

If a number plus its reciprocal equals 10/3, find the number.

Answer

Let the number be x. Then its reciprocal is 1/x, and x ≠ 0.

x + 1/x = 10/3
Multiply throughout by 3x:
3x² + 3 = 10x
3x² − 10x + 3 = 0

Split the middle term: two numbers with product 3 × 3 = 9 and sum −10 are −1 and −9.

3x² − 9x − x + 3 = 0
3x(x − 3) − 1(x − 3) = 0
(x − 3)(3x − 1) = 0
x = 3 or x = 1/3

Both answers are genuine, and they are reciprocals of each other.

Check: 3 + 1/3 = 10/3 ✓ 1/3 + 3 = 10/3 ✓

Why two answers had to appear: the condition x + 1/x = 10/3 does not distinguish between x and 1/x — swapping them leaves the left side unchanged. So the solutions must come in reciprocal pairs, and here that pair is 3 and 1/3. Notice their product is 1, which matches the constant term 3 divided by the leading coefficient 3.

Q9.

A rectangular pool has area 2x² + 7x + 3 square hastas. If its width is 2x + 1 hastas, find its length. Hasta was a unit used to measure length.

Answer

Length = area ÷ width, so factor the area and see which factor is the given width.

2x² + 7x + 3
Split 7x using two numbers with product 2 × 3 = 6 and sum 7: they are 1 and 6
= 2x² + x + 6x + 3
= x(2x + 1) + 3(2x + 1)
= (2x + 1)(x + 3)

Width = (2x + 1) hastas ⇒ Length = (x + 3) hastas

Did you know? The hasta (from the Sanskrit for ‘hand’) was a standard length in ancient India, measured from the elbow to the tip of the middle finger — roughly 45 cm. It appears throughout the Śulba Sūtras, where altar areas are prescribed in square hastas.

Q10.

*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, the expression must be zero at x = 2; likewise it must be zero at x = 1/2.

At x = 2: p(2)² + 5(2) + r = 0 ⇒ 4p + r = −10 …(1)

At x = 1/2: p(1/2)² + 5(1/2) + r = 0 ⇒ p/4 + r = −5/2
Multiply by 4: p + 4r = −10 …(2)

Subtract (2) from (1):

(4p + r) − (p + 4r) = −10 − (−10)
3p − 3r = 0
p = r ■

The two equations even pin down the values: putting r = p in (1) gives 5p = −10, so p = r = −2 and the expression is −2x² + 5x − 2 = −(2x − 1)(x − 2). Both stated factors are visible.

Why ‘factor’ means ‘root’: if x − k is a factor, then px² + 5x + r = (x − k) × (something). Substituting x = k makes the first bracket zero, so the whole product is zero. This is the Factor Theorem, and it converts a statement about factors into two ordinary equations you can solve.

Tip: Notice how the symmetry does the work. The two roots, 2 and 1/2, are reciprocals, and for a quadratic the product of the roots is (constant)/(leading coefficient) = r/p. Reciprocal roots multiply to 1, so r/p = 1 — the same conclusion in one line.

Q11.

*If a + b + c = 5 and ab + bc + ca = 10, then prove that a³ + b³ + c³ − 3abc = − 25.

Answer

Use the three-cubes identity from Section 4.7:

a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

The first bracket is given. For the second, first get a² + b² + c² from the square of the sum:

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
5² = a² + b² + c² + 2(10)
25 = a² + b² + c² + 20
a² + b² + c² = 5

Now substitute both pieces:

a³ + b³ + c³ − 3abc = (5)(5 − 10)
= 5 × (−5)
= −25 ■

Why no individual value of a, b or c is needed: the identity expresses the answer entirely in terms of the two symmetric quantities a + b + c and ab + bc + ca, and both are given. Symmetric expressions in three letters can almost always be rebuilt from these building blocks — that is what makes identities such powerful shortcuts.

Did you know? The numbers here are unusual: a² + b² + c² = 5 while ab + bc + ca = 10 forces a, b, c to be non-real. The proof does not care — an identity holds for all values, and only the two given sums were ever used.

Q12.

*By factoring the expression, check that n³ − n is always divisible by 6 for all natural numbers n. Give reasons.

Answer

Factor first.

n³ − n = n(n² − 1)
= n(n − 1)(n + 1)
= (n − 1) · n · (n + 1)

So n³ − n is the product of three consecutive natural numbers. Now argue divisibility.

Divisible by 2. Among any two consecutive integers one is even, so among three there is certainly an even one. Hence 2 divides the product.
Divisible by 3. Any integer leaves remainder 0, 1 or 2 on division by 3. Among three consecutive integers, all three remainders occur exactly once, so one of them is a multiple of 3. Hence 3 divides the product.
Therefore divisible by 6. The product is divisible by 2 and by 3, and 2 and 3 have no common factor other than 1, so their product 6 divides it.

n	(n − 1)n(n + 1)	n³ − n	÷ 6
2	1 × 2 × 3	6	1
5	4 × 5 × 6	120	20
10	9 × 10 × 11	990	165

Why ‘2 divides it and 3 divides it’ gives 6: this step needs 2 and 3 to be coprime. If a number is divisible by 4 and by 6 you may not conclude it is divisible by 24 — 12 is a counter-example. Here gcd(2, 3) = 1, so the conclusion is safe.

Try This: The same argument shows n⁵ − n is always divisible by 30, since n⁵ − n = (n − 1)n(n + 1)(n² + 1) and one can show 5 always divides it too.

Q13.

*Find the value of (i) x³ + y³ − 12xy + 64, when x + y = − 4 (ii) x³ − 8y³ − 36xy − 216, when x = 2y + 6

Answer

Both parts are the identity x³ + y³ + z³ − 3xyz in disguise, with the third letter chosen to make the first bracket vanish.

(i) x³ + y³ − 12xy + 64, when x + y = −4

Notice 64 = 4³ and 12xy = 3(x)(y)(4). Take z = 4:
x³ + y³ + 4³ − 3(x)(y)(4) = x³ + y³ + 64 − 12xy ✓

= (x + y + 4)(x² + y² + 16 − xy − 4y − 4x)
Given x + y = −4, the first bracket is (−4) + 4 = 0
Value = 0

(ii) x³ − 8y³ − 36xy − 216, when x = 2y + 6

Write −8y³ = (−2y)³ and −216 = (−6)³. Then with the three letters x, −2y, −6:
−3(x)(−2y)(−6) = −36xy ✓

x³ + (−2y)³ + (−6)³ − 3(x)(−2y)(−6)
= (x − 2y − 6)(x² + 4y² + 36 + 2xy − 12y + 6x)
Given x = 2y + 6, the first bracket is (2y + 6) − 2y − 6 = 0
Value = 0

Why both answers are zero: the identity says the expression equals (sum of the three letters) × (a second bracket). The condition supplied in each part is precisely the statement that the sum of the three letters is zero. So the whole product collapses, whatever the second bracket happens to be — you never need to evaluate it. Recognising the shape is the entire problem.

Tip: The general fact behind this is worth remembering: if x + y + z = 0 then x³ + y³ + z³ = 3xyz. Both parts above are that statement read backwards.

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