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

NCERT Class 9 Mathematics — Ganita Manjari Part I, Exploring Algebraic Identities, End-of-Chapter Exercises. All 13 questions solved.

This is the chapter’s full toolkit in one place. Worth having written out before starting:

$$(a \pm b)^2 = a^2 \pm 2ab + b^2 \qquad (a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca$$ $$a^2 – b^2 = (a+b)(a-b) \qquad (a \pm b)^3 = a^3 \pm 3a^2b + 3ab^2 \pm b^3$$ $$a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$$ $$a^3 + b^3 + c^3 – 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$

The last one is the workhorse of Questions 11 and 13, and it carries a consequence worth memorising: if $a + b + c = 0$ then $a^3 + b^3 + c^3 = 3abc$.

Questions $10$ to $13$ carry an asterisk, marking them as enrichment that is not assessed. All four are solved in full.

Key insight. Most of the “hard” questions here are one identity in disguise. Question 11 needs $a^2+b^2+c^2$, which comes from squaring $a+b+c$. Question 13 hides a third cube inside a constant — $64 = 4^3$ and $216 = 6^3$. Looking for a cube in a bare number is the single most useful habit in this exercise.

Question 1

Use suitable identities to find the following products:

Solution.

(i) $(-3x + 4)^2$

Take $a = -3x$, $b = 4$:

$$(-3x)^2 + 2(-3x)(4) + 4^2 = 9x^2 – 24x + 16$$

(ii) $(2s + 7)(2s – 7)$

Difference of squares, $a = 2s$, $b = 7$:

$$(2s)^2 – 7^2 = 4s^2 – 49$$

(iii) $\left( p^2 + \dfrac{1}{2} \right)\left( p^2 – \dfrac{1}{2} \right)$

$$\left(p^2\right)^2 – \left(\frac{1}{2}\right)^2 = p^4 – \frac{1}{4}$$

(iv) $(2n + 7)(2n – 7)$

$$4n^2 – 49$$

(v) $(s – 2t)(s^2 + 2st + 4t^2)$

This is $a^3 – b^3 = (a-b)(a^2+ab+b^2)$ with $a = s$, $b = 2t$ — check: $ab = 2st$ ✓ and $b^2 = 4t^2$ ✓

$$= s^3 – (2t)^3 = s^3 – 8t^3$$

(vi) $\left( \dfrac{1}{2r} – 4r \right)^2$

$$\left(\frac{1}{2r}\right)^2 – 2\left(\frac{1}{2r}\right)(4r) + (4r)^2 = \frac{1}{4r^2} – 4 + 16r^2$$

(the middle term’s $r$s cancel).

(vii) $(-3m + 4k – l)^2$

Three-term identity with $a = -3m$, $b = 4k$, $c = -l$:

$$9m^2 + 16k^2 + l^2 + 2(-3m)(4k) + 2(4k)(-l) + 2(-l)(-3m)$$ $$= 9m^2 + 16k^2 + l^2 – 24km – 8kl + 6lm$$

(viii) $\left( x – \dfrac{1}{3}y \right)^3$

Cube identity with $a = x$, $b = \frac{1}{3}y$:

$$x^3 – 3x^2\left(\frac{y}{3}\right) + 3x\left(\frac{y}{3}\right)^2 – \left(\frac{y}{3}\right)^3$$ $$= x^3 – x^2y + \frac{xy^2}{3} – \frac{y^3}{27}$$

(ix) $\left( \dfrac{7}{2}k – \dfrac{2}{3}m \right)^3$

Same identity with $a = \frac{7}{2}k$, $b = \frac{2}{3}m$:

$$\frac{343}{8}k^3 – 3\left(\frac{49}{4}k^2\right)\left(\frac{2}{3}m\right) + 3\left(\frac{7}{2}k\right)\left(\frac{4}{9}m^2\right) – \frac{8}{27}m^3$$ $$= \frac{343}{8}k^3 – \frac{49}{2}k^2m + \frac{14}{3}km^2 – \frac{8}{27}m^3$$

Question 2

Find the values using suitable identities:

Solution. Parts (i) to (iii) are products; (iv) to (viii) are cubes, using $(a \pm b)^3$.

(i) $17 \times 21$

Not symmetric about a round number, so use $(x+a)(x+b)$ with $x = 20$:

$$(20 – 3)(20 + 1) = 400 + 20(1 – 3) – 3 = 400 – 40 – 3 = 357$$

(ii) $104 \times 96$

Symmetric about $100$: use $a^2 – b^2$.

$$(100 + 4)(100 – 4) = 10000 – 16 = 9984$$

(iii) $24 \times 16$

Symmetric about $20$:

$$(20 + 4)(20 – 4) = 400 – 16 = 384$$

(iv) $147^3$

$147 = 150 – 3$, so $(a-b)^3 = a^3 – 3a^2b + 3ab^2 – b^3$:

$$150^3 – 3(150)^2(3) + 3(150)(9) – 27$$ $$= 3375000 – 202500 + 4050 – 27 = 3176523$$

(v) $199^3$

$199 = 200 – 1$:

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

(vi) $127^3$

$127 = 130 – 3$:

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

(vii) $(-107)^3$

A negative cubed is negative, so compute $107^3$ and attach a minus. With $107 = 100 + 7$:

$$1000000 + 3(10000)(7) + 3(100)(49) + 343 = 1000000 + 210000 + 14700 + 343 = 1225043$$

$$(-107)^3 = -1225043$$

(viii) $(-299)^3$

$299 = 300 – 1$:

$$27000000 – 3(90000)(1) + 3(300)(1) – 1 = 27000000 – 270000 + 900 – 1 = 26730899$$

$$(-299)^3 = -26730899$$

Question 3

Factor the following algebraic expressions:

Solution.

(i) $4y^2 + 1 + \dfrac{1}{16y^2}$

Three terms; the outer two are squares of $2y$ and $\frac{1}{4y}$, and the middle should be twice their product: $2(2y)\left(\frac{1}{4y}\right) = 1$ ✓

$$= \left( 2y + \frac{1}{4y} \right)^2$$

(ii) $9m^2 – \dfrac{1}{25n^2}$

Difference of squares:

$$= \left( 3m + \frac{1}{5n} \right)\left( 3m – \frac{1}{5n} \right)$$

(iii) $27b^3 – \dfrac{1}{64b^3}$

Difference of cubes, with $a = 3b$ and $b’ = \frac{1}{4b}$:

$$= \left( 3b – \frac{1}{4b} \right)\left( 9b^2 + \frac{3}{4} + \frac{1}{16b^2} \right)$$

(iv) $x^2 + \dfrac{5x}{6} + \dfrac{1}{6}$

Clear the fractions by taking out $\frac{1}{6}$:

$$= \frac{1}{6}\left( 6x^2 + 5x + 1 \right)$$

Split the middle term: two numbers multiplying to $6$ and adding to $5$ are $2$ and $3$.

$$6x^2 + 2x + 3x + 1 = 2x(3x+1) + (3x+1) = (3x+1)(2x+1)$$

$$= \frac{1}{6}(3x + 1)(2x + 1)$$

(v) $27u^3 – \dfrac{1}{125} – \dfrac{27u^2}{5} + \dfrac{9u}{25}$

Four terms in descending powers of $u$ — try the cube identity $(a-b)^3 = a^3 – 3a^2b + 3ab^2 – b^3$ with $a = 3u$, $b = \frac{1}{5}$:

$$a^3 = 27u^3 \; \checkmark \qquad -3a^2b = -3(9u^2)\left(\tfrac{1}{5}\right) = -\tfrac{27u^2}{5} \; \checkmark$$ $$3ab^2 = 3(3u)\left(\tfrac{1}{25}\right) = \tfrac{9u}{25} \; \checkmark \qquad -b^3 = -\tfrac{1}{125} \; \checkmark$$

$$= \left( 3u – \frac{1}{5} \right)^3$$

(vi) $64y^3 + \dfrac{1}{125}z^3$

Sum of cubes with $a = 4y$, $b = \frac{z}{5}$:

$$= \left( 4y + \frac{z}{5} \right)\left( 16y^2 – \frac{4yz}{5} + \frac{z^2}{25} \right)$$

(vii) $p^3 + 27q^3 + r^3 – 9pqr$

This is $a^3+b^3+c^3-3abc$ with $a = p$, $b = 3q$, $c = r$ — check $3abc = 3(p)(3q)(r) = 9pqr$ ✓

$$= (p + 3q + r)\left( p^2 + 9q^2 + r^2 – 3pq – 3qr – pr \right)$$

(viii) $9m^2 – 12m + 4$

A perfect square with $3m$ and $2$, since $2(3m)(2) = 12m$:

$$= (3m – 2)^2$$

(ix) $9x^3 – \dfrac{8}{3}y^3 + \dfrac{z^3}{3} + 6xyz$

Take out $\frac{1}{3}$ to clear the fractions:

$$= \frac{1}{3}\left( 27x^3 – 8y^3 + z^3 + 18xyz \right)$$

The bracket is $a^3+b^3+c^3-3abc$ with $a = 3x$, $b = -2y$, $c = z$: check $-3abc = -3(3x)(-2y)(z) = +18xyz$ ✓

$$= \frac{1}{3}(3x – 2y + z)\left( 9x^2 + 4y^2 + z^2 + 6xy + 2yz – 3xz \right)$$

(x) $4x^2 + 9y^2 + 36z^2 + 12xz + 36yz + 24xy$ — a misprint

As printed this expression cannot be factored at all. For a six-term expression to be $(a+b+c)^2$ the three cross terms must match the three squares. Here the squares are $(2x)^2$, $(3y)^2$ and $(6z)^2$, which force

$$2ab = 2(2x)(3y) = 12xy, \qquad 2bc = 2(3y)(6z) = 36yz, \qquad 2ca = 2(6z)(2x) = 24xz$$

The book prints $12xz$ and $24xy$ — the coefficients of $xy$ and $xz$ the wrong way round. The $36yz$ is right.

This is not merely a failure to be a perfect square: the expression as printed does not factor into two linear expressions at all, whatever they might be. (Writing it as a quadratic form, its $3 \times 3$ matrix has determinant $-2916$ and full rank; a product of two linear factors always gives rank at most $2$.)

With the two coefficients exchanged the expression becomes

$$4x^2 + 9y^2 + 36z^2 + 12xy + 36yz + 24xz = (2x + 3y + 6z)^2$$

which is certainly what was intended, and which matches the pattern of every other part of this question.

(xi) $27u^3 – \dfrac{1}{216} – \dfrac{9u^2}{2} + \dfrac{u}{4}$

Same shape as part (v), with $a = 3u$ and $b = \frac{1}{6}$:

$$a^3 = 27u^3 \; \checkmark \qquad -3a^2b = -3(9u^2)\left(\tfrac{1}{6}\right) = -\tfrac{9u^2}{2} \; \checkmark$$ $$3ab^2 = 3(3u)\left(\tfrac{1}{36}\right) = \tfrac{u}{4} \; \checkmark \qquad -b^3 = -\tfrac{1}{216} \; \checkmark$$

$$= \left( 3u – \frac{1}{6} \right)^3$$

Question 4

Simplify the following:

Solution. Factorise top and bottom, then cancel.

(i) $\dfrac{4x^2 + 4x + 1}{4x^2 – 1}$

$$\frac{(2x+1)^2}{(2x+1)(2x-1)} = \frac{2x+1}{2x-1}$$

(ii) $\dfrac{9(3a^3 – 24b^3)}{9a^2 – 36b^2}$

The numerator has a common factor of $3$ inside, and then a difference of cubes:

$$9 \times 3(a^3 – 8b^3) = 27(a – 2b)(a^2 + 2ab + 4b^2)$$

The denominator is $9(a^2 – 4b^2) = 9(a – 2b)(a + 2b)$. Cancelling $(a-2b)$ and dividing $27$ by $9$:

$$= \frac{3(a^2 + 2ab + 4b^2)}{a + 2b}$$

(iii) $\dfrac{s^3 + 125t^3}{s^2 – 2st – 35t^2}$

Numerator is a sum of cubes with $s$ and $5t$; denominator splits with $-7t$ and $5t$:

$$\frac{(s + 5t)(s^2 – 5st + 25t^2)}{(s – 7t)(s + 5t)} = \frac{s^2 – 5st + 25t^2}{s – 7t}$$

Question 5

Find possible expressions for the length and breadth of each of the following rectangles whose areas are given.

Solution. Area $=$ length $\times$ breadth, so factorising the area into two brackets gives a possible pair of dimensions.

(i) $25a^2 – 30ab + 9b^2$

A perfect square with $5a$ and $3b$, since $2(5a)(3b) = 30ab$:

$$= (5a – 3b)^2 = (5a – 3b)(5a – 3b)$$

So both sides are $(5a – 3b)$ — the rectangle is a square.

(ii) $36s^2 – 49t^2$

Difference of squares:

$$= (6s + 7t)(6s – 7t)$$

Question 6

Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given.

Solution. Volume $=$ length $\times$ breadth $\times$ height, so factorise into three brackets.

(i) $6a^2 – 24b^2$

Take out the common factor $6$, then a difference of squares:

$$6(a^2 – 4b^2) = 6(a – 2b)(a + 2b)$$

Three factors: $6$, $(a – 2b)$ and $(a + 2b)$.

(ii) $3ps^2 – 15ps + 12p$

Common factor $3p$, then a quadratic:

$$3p(s^2 – 5s + 4) = 3p(s – 1)(s – 4)$$

Three factors: $3p$, $(s – 1)$ and $(s – 4)$.

Question 7

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 $s$.

Solution. The path surrounds the playground, so the outer boundary is a larger square. The path adds $s$ metres on both sides, so the outer side is

$$40 + s + s = 40 + 2s$$

The path is what is left when the playground is removed from the larger square:

$$\text{area of path} = (40 + 2s)^2 – 40^2$$

Expand the first square:

$$(40 + 2s)^2 = 1600 + 2(40)(2s) + 4s^2 = 1600 + 160s + 4s^2$$

Subtract:

$$\text{area of path} = 1600 + 160s + 4s^2 – 1600 = 4s^2 + 160s$$

Check at $s = 0$: no path, and the formula gives $0$ ✓. At $s = 1$ it gives $164$ m², which is $42^2 – 40^2 = 1764 – 1600$ ✓

Question 8

If a number plus its reciprocal equals $\dfrac{10}{3}$, find the number.

Solution. Let the number be $x$; its reciprocal is $\frac{1}{x}$.

$$x + \frac{1}{x} = \frac{10}{3}$$

Multiply through by $3x$ to clear both denominators (valid since $x \neq 0$ — a number with a reciprocal cannot be zero):

$$3x^2 + 3 = 10x$$ $$3x^2 – 10x + 3 = 0$$

Split the middle term: two numbers multiplying to $3 \times 3 = 9$ and adding to $-10$, namely $-1$ and $-9$:

$$3x^2 – 9x – x + 3 = 3x(x – 3) – (x – 3) = (x – 3)(3x – 1) = 0$$

$$x = 3 \qquad \text{or} \qquad x = \frac{1}{3}$$

Both are genuine answers, and they are reciprocals of each other — which they must be, since the condition is symmetric between $x$ and $\frac{1}{x}$.

Check: $3 + \frac{1}{3} = \frac{10}{3}$ ✓ and $\frac{1}{3} + 3 = \frac{10}{3}$ ✓

Question 9

A rectangular pool has area $2x^2 + 7x + 3$ square hastas. If its width is $2x + 1$ hastas, find its length.

Solution. Length $=$ area $\div$ width, so factorise the area and cancel the known side.

Split the middle term of $2x^2 + 7x + 3$: two numbers multiplying to $2 \times 3 = 6$ and adding to $7$, namely $1$ and $6$:

$$2x^2 + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)$$

So

$$\text{length} = \frac{(2x+1)(x+3)}{2x+1} = x + 3$$

Check: $(2x+1)(x+3) = 2x^2 + 6x + x + 3 = 2x^2 + 7x + 3$ ✓

Question 10

*If both $x – 2$ and $x – \dfrac{1}{2}$ are factors of $px^2 + 5x + r$, show that $p = r$.

Solution. If $x – k$ is a factor of a polynomial, then substituting $x = k$ makes the polynomial zero. Use both factors.

From $x – 2$: substitute $x = 2$.

$$4p + 10 + r = 0 \qquad \text{(i)}$$

From $x – \frac{1}{2}$: substitute $x = \frac{1}{2}$.

$$\frac{p}{4} + \frac{5}{2} + r = 0$$

Multiply through by $4$ to clear the fraction:

$$p + 10 + 4r = 0 \qquad \text{(ii)}$$

Subtract (ii) from (i):

$$(4p + 10 + r) – (p + 10 + 4r) = 0$$ $$3p – 3r = 0$$ $$p = r$$

which is what was to be shown.

As a bonus, substituting $p = r$ into (i) gives $5p = -10$, so $p = r = -2$ and the polynomial is $-2x^2 + 5x – 2 = -(2x-1)(x-2)$ — which indeed has both stated factors.

Question 11

*If $a + b + c = 5$ and $ab + bc + ca = 10$, then prove that $a^3 + b^3 + c^3 – 3abc = -25$.

Solution. The identity to use is

$$a^3 + b^3 + c^3 – 3abc = (a+b+c)\left( a^2 + b^2 + c^2 – ab – bc – ca \right)$$

The first bracket is given as $5$. The second needs $a^2 + b^2 + c^2$, which is not given directly — but it comes from squaring the first condition:

$$(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$$ $$5^2 = a^2 + b^2 + c^2 + 2(10)$$ $$25 = a^2 + b^2 + c^2 + 20$$ $$a^2 + b^2 + c^2 = 5$$

Now substitute both into the identity:

$$a^3 + b^3 + c^3 – 3abc = 5\left( 5 – 10 \right) = 5(-5) = -25$$

as required.

Question 12

*By factoring the expression, check that $n^3 – n$ is always divisible by $6$ for all natural numbers $n$. Give reasons.

Solution. Factorise:

$$n^3 – n = n(n^2 – 1) = n(n – 1)(n + 1)$$

Reordering, this is

$$(n-1) \times n \times (n+1)$$

— the product of three consecutive integers. That is the whole answer, and the reasons follow from it.

Divisible by $2$. Among any two consecutive integers one is even, so among three there is certainly at least one even number. The product therefore has a factor of $2$.

Divisible by $3$. Among any three consecutive integers exactly one is a multiple of $3$ — whatever the remainder of $n-1$ on division by $3$, one of the three consecutive values hits $0$. So the product has a factor of $3$.

Therefore divisible by $6$. The product is divisible by both $2$ and $3$, and since $2$ and $3$ share no common factor, it is divisible by $2 \times 3 = 6$.

Check a few: $n = 2$ gives $6$; $n = 3$ gives $24$; $n = 4$ gives $60$; $n = 5$ gives $120$ — all multiples of $6$ ✓

Question 13

*Find the value of:

Solution. Both parts hide a third cube inside a constant, and both then use: if $a + b + c = 0$ then $a^3 + b^3 + c^3 = 3abc$, so $a^3+b^3+c^3-3abc = 0$.

(i) $x^3 + y^3 – 12xy + 64$, when $x + y = -4$

Notice $64 = 4^3$. Take $a = x$, $b = y$, $c = 4$; then

$$a^3+b^3+c^3-3abc = x^3 + y^3 + 64 – 3(x)(y)(4) = x^3 + y^3 + 64 – 12xy$$

which is exactly the expression. Now

$$a + b + c = x + y + 4 = -4 + 4 = 0$$

Since the sum is zero, the whole expression is zero.

$$x^3 + y^3 – 12xy + 64 = 0$$

(ii) $x^3 – 8y^3 – 36xy – 216$, when $x = 2y + 6$

Here $-8y^3 = (-2y)^3$ and $-216 = (-6)^3$. Take $a = x$, $b = -2y$, $c = -6$:

$$a^3+b^3+c^3-3abc = x^3 – 8y^3 – 216 – 3(x)(-2y)(-6) = x^3 – 8y^3 – 216 – 36xy$$

which is the expression. And

$$a + b + c = x – 2y – 6 = (2y + 6) – 2y – 6 = 0$$

So again the expression is zero.

$$x^3 – 8y^3 – 36xy – 216 = 0$$

Common mistakes

• Question 1(i), writing $(-3x)^2 = -9x^2$. Squaring a negative gives a positive: $9x^2$.
• Question 1(v), multiplying out term by term. It is $a^3 – b^3$ in factored form, so the answer is two terms, not six. Recognising the pattern saves all the work.
• Question 1(viii) and (ix), forgetting the middle coefficients. A cube has four terms, with coefficients $1, 3, 3, 1$. Writing only $a^3 – b^3$ loses two of them.
• Question 2(i), using $a^2 – b^2$. $17$ and $21$ are not equidistant from a round number ($17 = 20-3$ but $21 = 20+1$), so the difference-of-squares identity does not apply.
• Question 2(vii), forgetting the sign. An odd power of a negative number is negative: $(-107)^3 = -1{,}225{,}043$.
• Question 3(iv) and (ix), leaving fractions inside. Taking out $\frac{1}{6}$ or $\frac{1}{3}$ first turns awkward coefficients into whole numbers and makes the pattern visible.
• Question 3(x), trying to force a factorisation. As printed the expression genuinely does not factor. The $xy$ and $xz$ coefficients have been exchanged; with them the right way round it is $(2x + 3y + 6z)^2$. Checking the three cross terms against the three squares is what exposes this.
• Question 5(i), giving different length and breadth. $25a^2 – 30ab + 9b^2$ is a perfect square, so both sides are $5a – 3b$ and the rectangle is a square. That is a valid answer.
• Question 6, stopping at two factors. A cuboid has three dimensions, so the volume must be split into three brackets — the numerical common factor counts as one of them.
• Question 7, using $40 + s$ for the outer side. The path runs all the way round, adding $s$ on each of two opposite sides, so the outer side is $40 + 2s$. Using $40 + s$ gives $s^2 + 80s$, about half the true area.
• Question 8, giving only $x = 3$. The quadratic has two roots and both satisfy the condition. They are reciprocals, which is the check that neither has been lost.
• Question 10, substituting $x = -2$ and $x = -\frac{1}{2}$. The factor $x – 2$ is zero at $x = +2$, not $-2$. Getting the sign wrong here gives $p = -r$.
• Question 11, trying to find $a$, $b$ and $c$ individually. They cannot be determined — and need not be. Only the two symmetric combinations are needed, and squaring the first condition supplies the missing one.
• Question 13, missing the hidden cube. $64$ and $216$ look like ordinary constants until you notice $64 = 4^3$ and $216 = 6^3$. Without spotting that, there is no way into either part.

Practise next

• Chapter 5, I’m Up and Down, and Round and Round — the next chapter, which turns from algebra to the geometry of circles.
• Exercise Set 4.5 — worth returning to after Question 4 here, since both rest on factorising sums and differences of cubes.
• Exercise Set 4.3, Question 2 — six-term perfect squares, the skill Question 3(x) here depends on and which is what reveals the misprint.