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NCERT Solutions · Class 9 Maths · Ganita Manjari Part I · Chapter 4

Chapter 4: Exploring Algebraic Identities (Algebraic Identities)

Step-by-step solutions to Exercise Sets 4.1 to 4.5 and all End-of-Chapter Exercises of Ganita Manjari (NCERT Class 9 Maths, 2026-27): expanding squares and cubes, factorisation using identities, splitting the middle term, and simplifying rational expressions. All 25 questions are answered, with the key answer highlighted.

Identities used in this chapter: (1) $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ (2) $(a - b)^{2} = a^{2} - 2 ab + b^{2}$ (3) $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 ab + 2 b c + 2 c a$ (4) $a^{2} - b^{2} = (a + b) (a - b)$ (5) $(x + a) (x + b) = x^{2} + (a + b) x + ab$ (6) $(a \pm b)^{3} = a^{3} \pm 3 a^{2} b + 3 a b^{2} \pm b^{3}$ (7) $a^{3} \pm b^{3} = (a \pm b) (a^{2} \mp ab + b^{2})$ (8) $a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a)$ .

Exercise Set 4.1

Using the identity

(a + b)^{2} = a^{2} + 2 ab + b^{2}

, expand the following: (i)

(7 x + 4 y)^{2}

(ii)

(\frac{7}{5} x + \frac{3}{2} y)^{2}

(iii)

(2.5 p + 1.5 q)^{2}

(iv)

(\frac{3}{4} s + 8 t)^{2}

(v)

(x + \frac{1}{2 y})^{2}

(vi)

(\frac{1}{x} + \frac{1}{y})^{2}

Solution

(i) $a = 7 x$ , $b = 4 y$ :

$(7 x)^{2} + 2 (7 x) (4 y) + (4 y)^{2} = 49 x^{2} + 56 x y + 16 y^{2}$

(ii) $a = \frac{7}{5} x$ , $b = \frac{3}{2} y$ :

$\frac{49}{25} x^{2} + 2 \cdot \frac{7}{5} \cdot \frac{3}{2} x y + \frac{9}{4} y^{2} = \frac{49}{25} x^{2} + \frac{21}{5} x y + \frac{9}{4} y^{2}$

(iii) $a = 2.5 p$ , $b = 1.5 q$ :

$6.25 p^{2} + 2 (2.5) (1.5) pq + 2.25 q^{2} = 6.25 p^{2} + 7.5 pq + 2.25 q^{2}$

(iv) $a = \frac{3}{4} s$ , $b = 8 t$ :

$\frac{9}{16} s^{2} + 2 \cdot \frac{3}{4} \cdot 8 s t + 64 t^{2} = \frac{9}{16} s^{2} + 12 s t + 64 t^{2}$

(v) $a = x$ , $b = \frac{1}{2 y}$ :

$x^{2} + 2 \cdot x \cdot \frac{1}{2 y} + \frac{1}{4 y ^{2}} = x^{2} + \frac{x}{y} + \frac{1}{4 y ^{2}}$

(vi) $a = \frac{1}{x}$ , $b = \frac{1}{y}$ :

$\frac{1}{x ^{2}} + \frac{2}{x y} + \frac{1}{y ^{2}}$

(i) $49 x^{2} + 56 x y + 16 y^{2}$ (ii) $\frac{49}{25} x^{2} + \frac{21}{5} x y + \frac{9}{4} y^{2}$ (iii) $6.25 p^{2} + 7.5 pq + 2.25 q^{2}$ (iv) $\frac{9}{16} s^{2} + 12 s t + 64 t^{2}$ (v) $x^{2} + \frac{x}{y} + \frac{1}{4 y ^{2}}$ (vi) $\frac{1}{x ^{2}} + \frac{2}{x y} + \frac{1}{y ^{2}}$

Using the same identity, find the values of the following: (i)

(64)^{2}

(ii)

(105)^{2}

(iii)

(205)^{2}

Solution

$(i) (60 + 4)^{2} = 3600 + 2 (60) (4) + 16 = 3600 + 480 + 16 = 4096$

$(ii) (100 + 5)^{2} = 10000 + 1000 + 25 = 11025$

$(iii) (200 + 5)^{2} = 40000 + 2000 + 25 = 42025$

(i) 4096 (ii) 11025 (iii) 42025

Exercise Set 4.2

Factor completely: (i)

9 x^{2} + 24 x y + 16 y^{2}

(ii)

4 s^{2} + 20 s t + 25 t^{2}

(iii)

49 x^{2} + 28 x y + 4 y^{2}

(iv)

64 p^{2} + \frac{32}{3} pq + \frac{4}{9} q^{2}

(v)

3 a^{2} + 4 ab + \frac{4}{3} b^{2}

(vi)

\frac{9}{5} s^{2} + 6 s v + 5 v^{2}

Solution

In each case look for $a^{2} + 2 ab + b^{2}$ : find the two square terms and check that the middle term is twice their product.

(i) $9 x^{2} = (3 x)^{2}$ , $16 y^{2} = (4 y)^{2}$ , $2 (3 x) (4 y) = 24 x y$ ✓, so $9 x^{2} + 24 x y + 16 y^{2} = (3 x + 4 y)^{2}$ .

(ii) $(2 s)^{2} + 2 (2 s) (5 t) + (5 t)^{2} = (2 s + 5 t)^{2}$ .

(iii) $(7 x)^{2} + 2 (7 x) (2 y) + (2 y)^{2} = (7 x + 2 y)^{2}$ .

(iv) $64 p^{2} = (8 p)^{2}$ , $\frac{4}{9} q^{2} = (\frac{2}{3} q)^{2}$ , $2 (8 p) (\frac{2}{3} q) = \frac{32}{3} pq$ ✓, so the expression is $(8 p + \frac{2}{3} q)^{2}$ .

(v) Take $\frac{1}{3}$ out as a common factor (as 2 was taken out in Example 7):

$3 a^{2} + 4 ab + \frac{4}{3} b^{2} = \frac{1}{3} (9 a^{2} + 12 ab + 4 b^{2}) = \frac{1}{3} (3 a + 2 b)^{2}$

(vi) Take $\frac{1}{5}$ out:

$\frac{9}{5} s^{2} + 6 s v + 5 v^{2} = \frac{1}{5} (9 s^{2} + 30 s v + 25 v^{2}) = \frac{1}{5} (3 s + 5 v)^{2}$

(i) $(3 x + 4 y)^{2}$ (ii) $(2 s + 5 t)^{2}$ (iii) $(7 x + 2 y)^{2}$ (iv) $(8 p + \frac{2}{3} q)^{2}$ (v) $\frac{1}{3} (3 a + 2 b)^{2}$ (vi) $\frac{1}{5} (3 s + 5 v)^{2}$

Find the values of the following using the identity

(a - b)^{2} = a^{2} - 2 ab + b^{2}

: (i)

(79)^{2}

(ii)

(193)^{2}

(iii)

(299)^{2}

Solution

$(i) (80 - 1)^{2} = 6400 - 160 + 1 = 6241$

$(ii) (200 - 7)^{2} = 40000 - 2800 + 49 = 37249$

$(iii) (300 - 1)^{2} = 90000 - 600 + 1 = 89401$

(i) 6241 (ii) 37249 (iii) 89401

Exercise Set 4.3

Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier. (i)

11 7^{2}

(ii)

7 8^{2}

(iii)

19 8^{2}

(iv)

21 4^{2}

(v)

110 4^{2}

(vi)

112 0^{2}

Solution

Numbers just above a round number use $(a + b)^{2}$ (or $(a + b + c)^{2}$ ); numbers just below use $(a - b)^{2}$ .

$(i) 11 7^{2} = (100 + 10 + 7)^{2} = 10000 + 100 + 49 + 2000 + 1400 + 140 = 13689$

$(ii) 7 8^{2} = (80 - 2)^{2} = 6400 - 320 + 4 = 6084$

$(iii) 19 8^{2} = (200 - 2)^{2} = 40000 - 800 + 4 = 39204$

$(iv) 21 4^{2} = (200 + 10 + 4)^{2} = 40000 + 100 + 16 + 4000 + 1600 + 80 = 45796$

$(v) 110 4^{2} = (1100 + 4)^{2} = 1210000 + 8800 + 16 = 1218816$

$(vi) 112 0^{2} = (1100 + 20)^{2} = 1210000 + 44000 + 400 = 1254400$

(i) 13689 (ii) 6084 (iii) 39204 (iv) 45796 (v) 1218816 (vi) 1254400. $(a - b)^{2}$ is easiest for 78 and 198; $(a + b)^{2}$ or $(a + b + c)^{2}$ for the others.

Factor using suitable identities: (i)

16 y^{2} - 24 y + 9

(ii)

\frac{9}{4} s^{2} + 6 s t + 4 t^{2}

(iii)

\frac{m ^{2}}{9} + \frac{mk}{3} + \frac{k ^{2}}{4} + 3 nk + 2 mn + 9 n^{2}

(iv)

\frac{p ^{2}}{16} - 2 + \frac{16}{p ^{2}}

(v)

9 a^{2} + 4 b^{2} + c^{2} - 12 ab + 6 a c - 4 b c

Solution

(i) $(4 y)^{2} - 2 (4 y) (3) + 3^{2} = (4 y - 3)^{2}$ .

(ii) $(\frac{3}{2} s)^{2} + 2 (\frac{3}{2} s) (2 t) + (2 t)^{2} = (\frac{3}{2} s + 2 t)^{2}$ .

(iii) The three squares are $(\frac{m}{3})^{2}$ , $(\frac{k}{2})^{2}$ , $(3 n)^{2}$ . Check the products: $2 \cdot \frac{m}{3} \cdot \frac{k}{2} = \frac{mk}{3}$ ✓, $2 \cdot \frac{k}{2} \cdot 3 n = 3 nk$ ✓, $2 \cdot \frac{m}{3} \cdot 3 n = 2 mn$ ✓. So it is $(\frac{m}{3} + \frac{k}{2} + 3 n)^{2}$ .

(iv) $(\frac{p}{4})^{2} - 2 \cdot \frac{p}{4} \cdot \frac{4}{p} + (\frac{4}{p})^{2} = (\frac{p}{4} - \frac{4}{p})^{2}$ (since $2 \cdot \frac{p}{4} \cdot \frac{4}{p} = 2$ ).

(v) The squares are $(3 a)^{2}$ , $(2 b)^{2}$ , $c^{2}$ . The signs of the products tell us the signs: $- 12 ab$ means 3a and 2b have opposite signs; $+ 6 a c$ means 3a and c have the same sign; $- 4 b c$ means 2b and c have opposite signs. All three agree with $3 a$ , $- 2 b$ , $c$ :

$(3 a - 2 b + c)^{2} = 9 a^{2} + 4 b^{2} + c^{2} - 12 ab - 4 b c + 6 a c ✓$

(i) $(4 y - 3)^{2}$ (ii) $(\frac{3}{2} s + 2 t)^{2}$ (iii) $(\frac{m}{3} + \frac{k}{2} + 3 n)^{2}$ (iv) $(\frac{p}{4} - \frac{4}{p})^{2}$ (v) $(3 a - 2 b + c)^{2}$

Expand the following using the identity

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 ab + 2 b c + 2 c a

: (i)

(p + 3 q + 7 r)^{2}

(ii)

(3 x - 2 y + 4 z)^{2}

Solution

(i) With $a = p$ , $b = 3 q$ , $c = 7 r$ :

$p^{2} + 9 q^{2} + 49 r^{2} + 2 (p) (3 q) + 2 (3 q) (7 r) + 2 (7 r) (p) = p^{2} + 9 q^{2} + 49 r^{2} + 6 pq + 42 q r + 14 r p$

(ii) With $a = 3 x$ , $b = - 2 y$ , $c = 4 z$ :

$9 x^{2} + 4 y^{2} + 16 z^{2} + 2 (3 x) (- 2 y) + 2 (- 2 y) (4 z) + 2 (4 z) (3 x) = 9 x^{2} + 4 y^{2} + 16 z^{2} - 12 x y - 16 y z + 24 z x$

(i) $p^{2} + 9 q^{2} + 49 r^{2} + 6 pq + 42 q r + 14 r p$ (ii) $9 x^{2} + 4 y^{2} + 16 z^{2} - 12 x y - 16 y z + 24 z x$

Is this an identity?

(a + b - c)^{2} + (a - b + c)^{2} + (a - b - c)^{2} = 2 a^{2} + 2 b^{2} + 2 c^{2}

Solution

Expand each square using $(a + b + c)^{2}$ with the right signs:

$(a + b - c)^{2} = a^{2} + b^{2} + c^{2} + 2 ab - 2 b c - 2 c a$

$(a - b + c)^{2} = a^{2} + b^{2} + c^{2} - 2 ab - 2 b c + 2 c a$

$(a - b - c)^{2} = a^{2} + b^{2} + c^{2} - 2 ab + 2 b c - 2 c a$

Adding:

$LHS = 3 a^{2} + 3 b^{2} + 3 c^{2} - 2 ab - 2 b c - 2 c a$

This is not equal to $2 a^{2} + 2 b^{2} + 2 c^{2}$ for all values. For example, $a = b = c = 1$ gives LHS $= 1 + 1 + 1 = 3$ but RHS $= 6$ .

(It is true only for the particular values where $a^{2} + b^{2} + c^{2} = 2 (ab + b c + c a)$ , which is not all values.)

No, it is not an identity: the LHS equals $3 a^{2} + 3 b^{2} + 3 c^{2} - 2 ab - 2 b c - 2 c a$ (e.g. a = b = c = 1 gives 3 ≠ 6).

Exercise Set 4.4

Fill in the blanks to complete the following identities: (i)

s^{2} - 11 s + 24 = (__) (__)

(ii)

(__) (x + 1) = (3 x^{2} - 4 x - 7)

(iii)

10 x^{2} - 11 x - 6 = (2 x -__) (__+ 2)

(iv)

6 x^{2} + 7 x + 2 = (__) (__)

Solution

(i) We need two numbers with sum −11 and product 24: −3 and −8.

$s^{2} - 11 s + 24 = (s - 3) (s - 8)$

(ii) The missing factor times $(x + 1)$ must give $3 x^{2}$ and $- 7$ , so it is $(3 x - 7)$ . Check: $(3 x - 7) (x + 1) = 3 x^{2} + 3 x - 7 x - 7 = 3 x^{2} - 4 x - 7$ ✓

(iii) $(2 x - k) (m x + 2)$ : the $x^{2}$ term gives $2 m = 10$ , so $m = 5$ ; the constant gives $- 2 k = - 6$ , so $k = 3$ . Check: $(2 x - 3) (5 x + 2) = 10 x^{2} + 4 x - 15 x - 6 = 10 x^{2} - 11 x - 6$ ✓

(iv) Product of first and last coefficients: $6 \times 2 = 12$ . Split $7 x$ as $3 x + 4 x$ (since $3 \times 4 = 12$ , $3 + 4 = 7$ ):

$6 x^{2} + 3 x + 4 x + 2 = 3 x (2 x + 1) + 2 (2 x + 1) = (2 x + 1) (3 x + 2)$

(i) $(s - 3) (s - 8)$ (ii) $(3 x - 7)$ (iii) $(2 x - 3) (5 x + 2)$ (iv) $(2 x + 1) (3 x + 2)$

Select and use the identity that will help you to find the following products without multiplying directly: (i)

(41)^{2}

(ii)

(27)^{2}

(iii)

(23 \times 17)

(iv)

(135)^{2}

(v)

(97)^{2}

(vi)

(18 \times 29)

(vii)

(34 \times 43)

(viii)

(205)^{2}

Solution

$(i) (40 + 1)^{2} = 1600 + 80 + 1 = 1681$

$(ii) (30 - 3)^{2} = 900 - 180 + 9 = 729$

$(iii) (20 + 3) (20 - 3) = 2 0^{2} - 3^{2} = 400 - 9 = 391$

$(iv) (100 + 35)^{2} = 10000 + 7000 + 1225 = 18225$

$(v) (100 - 3)^{2} = 10000 - 600 + 9 = 9409$

(vi) Use $(x + a) (x + b) = x^{2} + (a + b) x + ab$ with $x = 20$ , $a = - 2$ , $b = 9$ :

$18 \times 29 = 400 + 7 \times 20 + (- 18) = 400 + 140 - 18 = 522$

(vii) With $x = 40$ , $a = - 6$ , $b = 3$ :

$34 \times 43 = 1600 + (- 3) (40) + (- 18) = 1600 - 120 - 18 = 1462$

$(viii) (200 + 5)^{2} = 40000 + 2000 + 25 = 42025$

(i) 1681 (ii) 729 (iii) 391 (iv) 18225 (v) 9409 (vi) 522 (vii) 1462 (viii) 42025

Factor the following: (i)

9 a^{2} + b^{2} + 4 c^{2} - 6 ab + 12 a c - 4 b c

(ii)

16 s^{2} + 25 t^{2} - 40 s t

(iii)

r^{2} - r - 42

(iv)

49 g^{2} + 14 g h + h^{2}

(v)

64 u^{2} + 121 v^{2} + 4 w^{2} - 176 uv - 32 u w + 44 v w

Solution

(i) Squares $(3 a)^{2}, b^{2}, (2 c)^{2}$ . Signs: $- 6 ab$ (3a, b opposite), $+ 12 a c$ (3a, 2c same), $- 4 b c$ (b, 2c opposite). So the terms are $3 a, - b, 2 c$ :

$(3 a - b + 2 c)^{2}$

(ii) $(4 s)^{2} - 2 (4 s) (5 t) + (5 t)^{2} = (4 s - 5 t)^{2}$ .

(iii) Two numbers with sum −1 and product −42: −7 and 6. So $r^{2} - r - 42 = (r - 7) (r + 6)$ .

(iv) $(7 g)^{2} + 2 (7 g) (h) + h^{2} = (7 g + h)^{2}$ .

(v) Squares $(8 u)^{2}, (11 v)^{2}, (2 w)^{2}$ . Signs: $- 176 uv$ (u, v opposite), $- 32 u w$ (u, w opposite), $+ 44 v w$ (v, w same). Terms $8 u, - 11 v, - 2 w$ . Check: $2 (8 u) (- 11 v) = - 176 uv$ ✓, $2 (8 u) (- 2 w) = - 32 u w$ ✓, $2 (- 11 v) (- 2 w) = 44 v w$ ✓.

$(8 u - 11 v - 2 w)^{2}$

(i) $(3 a - b + 2 c)^{2}$ (ii) $(4 s - 5 t)^{2}$ (iii) $(r - 7) (r + 6)$ (iv) $(7 g + h)^{2}$ (v) $(8 u - 11 v - 2 w)^{2}$

Exercise Set 4.5

Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero: (i)

\frac{3 p ^{2} - 3 pq - 18 q ^{2}}{p ^{2} + 3 pq - 10 q ^{2}}

(ii)

\frac{n ^{3} - 3 n ^{2} m + 3 n m ^{2} - m ^{3}}{5 m ^{2} - 10 mn + 5 n ^{2}}

(iii)

\frac{w ^{3} - v ^{3} + x ^{3} + 3 w v x}{w ^{2} + v ^{2} + x ^{2} - 2 w v - 2 v x + 2 w x}

(iv)

\frac{4 y ^{2} - 20 y z + 25 z ^{2}}{( 25 z ^{2} - 4 y ^{2} )}

(v)

\frac{( x ^{2} + x - 6 ) ( x ^{2} - 7 x + 12 )}{( x ^{2} - 6 x + 8 ) ( x ^{2} - 9 )}

(vi)

\frac{p ^{4} - 16}{p ^{2} - 4 p + 4}

Solution

Factor the numerator and the denominator, then cancel common factors.

(i) Numerator: $3 (p^{2} - pq - 6 q^{2}) = 3 (p - 3 q) (p + 2 q)$ . Denominator: $(p + 5 q) (p - 2 q)$ .

$\frac{3 ( p - 3 q ) ( p + 2 q )}{( p + 5 q ) ( p - 2 q )}$

There is no common factor, so this is already in simplest form.

(ii) Numerator: $(n - m)^{3}$ . Denominator: $5 (m^{2} - 2 mn + n^{2}) = 5 (n - m)^{2}$ .

$\frac{( n - m ) ^{3}}{5 ( n - m ) ^{2}} = \frac{n - m}{5}$

(iii) Numerator: with $a = w$ , $b = - v$ , $c = x$ , we have $a^{3} + b^{3} + c^{3} - 3 ab c = w^{3} - v^{3} + x^{3} + 3 w v x$ . So it equals $(w - v + x) (w^{2} + v^{2} + x^{2} + w v + v x - w x)$ . Denominator: $(w - v + x)^{2}$ .

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

(iv) Numerator: $(2 y - 5 z)^{2} = (5 z - 2 y)^{2}$ . Denominator: $(5 z - 2 y) (5 z + 2 y)$ .

$\frac{( 5 z - 2 y ) ^{2}}{( 5 z - 2 y ) ( 5 z + 2 y )} = \frac{5 z - 2 y}{5 z + 2 y}$

(v) $x^{2} + x - 6 = (x + 3) (x - 2)$ , $x^{2} - 7 x + 12 = (x - 3) (x - 4)$ , $x^{2} - 6 x + 8 = (x - 2) (x - 4)$ , $x^{2} - 9 = (x - 3) (x + 3)$ .

$\frac{( x + 3 ) ( x - 2 ) ( x - 3 ) ( x - 4 )}{( x - 2 ) ( x - 4 ) ( x - 3 ) ( x + 3 )} = 1$

(vi) $p^{4} - 16 = (p^{2} - 4) (p^{2} + 4) = (p - 2) (p + 2) (p^{2} + 4)$ and $p^{2} - 4 p + 4 = (p - 2)^{2}$ .

$\frac{( p - 2 ) ( p + 2 ) ( p ^{2} + 4 )}{( p - 2 ) ^{2}} = \frac{( p + 2 ) ( p ^{2} + 4 )}{p - 2}$

(i) $\frac{3 ( p - 3 q ) ( p + 2 q )}{( p + 5 q ) ( p - 2 q )}$ (ii) $\frac{n - m}{5}$ (iii) $\frac{w ^{2} + v ^{2} + x ^{2} + w v + v x - w x}{w - v + x}$ (iv) $\frac{5 z - 2 y}{5 z + 2 y}$ (v) 1 (vi) $\frac{( p + 2 ) ( p ^{2} + 4 )}{p - 2}$

End-of-Chapter Exercises

Use suitable identities to find the following products: (i)

(- 3 x + 4)^{2}

(ii)

(2 s + 7) (2 s - 7)

(iii)

(p^{2} + \frac{1}{2}) (p^{2} - \frac{1}{2})

(iv)

(2 n + 7) (2 n - 7)

(v)

(s - 2 t) (s^{2} + 2 s t + 4 t^{2})

(vi)

(\frac{1}{2 r} - 4 r)^{2}

(vii)

(- 3 m + 4 k - l)^{2}

(viii)

(x - \frac{1}{3} y)^{3}

(ix)

(\frac{7}{2} k - \frac{2}{3} m)^{3}

Solution

$(i) (4 - 3 x)^{2} = 16 - 24 x + 9 x^{2}$

$(ii) (2 s)^{2} - 7^{2} = 4 s^{2} - 49$

$(iii) (p^{2})^{2} - (\frac{1}{2})^{2} = p^{4} - \frac{1}{4}$

$(iv) 4 n^{2} - 49$

(v) This is $(a - b) (a^{2} + ab + b^{2}) = a^{3} - b^{3}$ with $a = s$ , $b = 2 t$ :

$s^{3} - 8 t^{3}$

$(vi) \frac{1}{4 r ^{2}} - 2 \cdot \frac{1}{2 r} \cdot 4 r + 16 r^{2} = \frac{1}{4 r ^{2}} - 4 + 16 r^{2}$

(vii) With $a = - 3 m$ , $b = 4 k$ , $c = - l$ :

$9 m^{2} + 16 k^{2} + l^{2} + 2 (- 3 m) (4 k) + 2 (4 k) (- l) + 2 (- l) (- 3 m) = 9 m^{2} + 16 k^{2} + l^{2} - 24 mk - 8 k l + 6 l m$

(viii) $(a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}$ with $a = x$ , $b = \frac{y}{3}$ :

$x^{3} - 3 x^{2} \cdot \frac{y}{3} + 3 x \cdot \frac{y ^{2}}{9} - \frac{y ^{3}}{27} = x^{3} - x^{2} y + \frac{x y ^{2}}{3} - \frac{y ^{3}}{27}$

(ix) With $a = \frac{7}{2} k$ , $b = \frac{2}{3} m$ :

$\frac{343}{8} k^{3} - 3 \cdot \frac{49}{4} k^{2} \cdot \frac{2}{3} m + 3 \cdot \frac{7}{2} k \cdot \frac{4}{9} m^{2} - \frac{8}{27} m^{3} = \frac{343}{8} k^{3} - \frac{49}{2} k^{2} m + \frac{14}{3} k m^{2} - \frac{8}{27} m^{3}$

(i) $9 x^{2} - 24 x + 16$ (ii) $4 s^{2} - 49$ (iii) $p^{4} - \frac{1}{4}$ (iv) $4 n^{2} - 49$ (v) $s^{3} - 8 t^{3}$ (vi) $\frac{1}{4 r ^{2}} - 4 + 16 r^{2}$ (vii) $9 m^{2} + 16 k^{2} + l^{2} - 24 mk - 8 k l + 6 l m$ (viii) $x^{3} - x^{2} y + \frac{x y ^{2}}{3} - \frac{y ^{3}}{27}$ (ix) $\frac{343}{8} k^{3} - \frac{49}{2} k^{2} m + \frac{14}{3} k m^{2} - \frac{8}{27} m^{3}$

Find the values using suitable identities: (i)

17 \times 21

(ii)

104 \times 96

(iii)

24 \times 16

(iv)

14 7^{3}

(v)

19 9^{3}

(vi)

12 7^{3}

(vii)

(- 107)^{3}

(viii)

(- 299)^{3}

Solution

Products of two numbers equally spaced around a round number use $(a + b) (a - b) = a^{2} - b^{2}$ :

$(i) (19 - 2) (19 + 2) = 361 - 4 = 357 (ii) (100 + 4) (100 - 4) = 10000 - 16 = 9984 (iii) (20 + 4) (20 - 4) = 400 - 16 = 384$

Cubes use $(a \pm b)^{3} = a^{3} \pm 3 a^{2} b + 3 a b^{2} \pm b^{3}$ :

$(iv) (150 - 3)^{3} = 3375000 - 3 (22500) (3) + 3 (150) (9) - 27 = 3375000 - 202500 + 4050 - 27 = 3176523$

$(v) (200 - 1)^{3} = 8000000 - 120000 + 600 - 1 = 7880599$

$(vi) (130 - 3)^{3} = 2197000 - 152100 + 3510 - 27 = 2048383$

$(vii) (- 107)^{3} = - (100 + 7)^{3} = - (1000000 + 210000 + 14700 + 343) = - 1225043$

$(viii) (- 299)^{3} = - (300 - 1)^{3} = - (27000000 - 270000 + 900 - 1) = - 26730899$

(i) 357 (ii) 9984 (iii) 384 (iv) 3176523 (v) 7880599 (vi) 2048383 (vii) −1225043 (viii) −26730899

Factor the following algebraic expressions: (i)

4 y^{2} + 1 + \frac{1}{16 y ^{2}}

(ii)

9 m^{2} - \frac{1}{25 n ^{2}}

(iii)

27 b^{3} - \frac{1}{64 b ^{3}}

(iv)

x^{2} + \frac{5 x}{6} + \frac{1}{6}

(v)

27 u^{3} - \frac{1}{125} - \frac{27 u ^{2}}{5} + \frac{9 u}{25}

(vi)

64 y^{3} + \frac{1}{125} z^{3}

(vii)

p^{3} + 27 q^{3} + r^{3} - 9 pq r

(viii)

9 m^{2} - 12 m + 4

(ix)

9 x^{3} - \frac{8}{3} y^{3} + \frac{z ^{3}}{3} + 6 x y z

(x)

4 x^{2} + 9 y^{2} + 36 z^{2} + 12 x z + 36 y z + 24 x y

(xi)

27 u^{3} - \frac{1}{216} - \frac{9 u ^{2}}{2} + \frac{u}{4}

Solution

(i) $(2 y)^{2} + 2 (2 y) (\frac{1}{4 y}) + (\frac{1}{4 y})^{2} = (2 y + \frac{1}{4 y})^{2}$ (middle term $= 1$ ✓).

(ii) Difference of squares: $(3 m)^{2} - (\frac{1}{5 n})^{2} = (3 m - \frac{1}{5 n}) (3 m + \frac{1}{5 n})$ .

(iii) Difference of cubes with $a = 3 b$ , $c = \frac{1}{4 b}$ :

$(3 b - \frac{1}{4 b}) (9 b^{2} + \frac{3}{4} + \frac{1}{16 b ^{2}})$

(iv) Need two numbers with sum $\frac{5}{6}$ and product $\frac{1}{6}$ : $\frac{1}{2}$ and $\frac{1}{3}$ . So $(x + \frac{1}{2}) (x + \frac{1}{3})$ .

(v) Rearranged: $27 u^{3} - \frac{27 u ^{2}}{5} + \frac{9 u}{25} - \frac{1}{125}$ . Compare with $(a - b)^{3}$ for $a = 3 u$ , $b = \frac{1}{5}$ : $3 a^{2} b = 3 \cdot 9 u^{2} \cdot \frac{1}{5} = \frac{27 u ^{2}}{5}$ ✓ and $3 a b^{2} = 3 \cdot 3 u \cdot \frac{1}{25} = \frac{9 u}{25}$ ✓. So it is $(3 u - \frac{1}{5})^{3}$ .

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

$(4 y + \frac{z}{5}) (16 y^{2} - \frac{4 y z}{5} + \frac{z ^{2}}{25})$

(vii) With $a = p$ , $b = 3 q$ , $c = r$ : $3 ab c = 9 pq r$ . So by identity (8):

$(p + 3 q + r) (p^{2} + 9 q^{2} + r^{2} - 3 pq - 3 q r - r p)$

(viii) $(3 m)^{2} - 2 (3 m) (2) + 2^{2} = (3 m - 2)^{2}$ .

(ix) Take $\frac{1}{3}$ out: $\frac{1}{3} (27 x^{3} - 8 y^{3} + z^{3} + 18 x y z)$ . With $a = 3 x$ , $b = - 2 y$ , $c = z$ : $a^{3} + b^{3} + c^{3} = 27 x^{3} - 8 y^{3} + z^{3}$ and $- 3 ab c = - 3 (3 x) (- 2 y) (z) = 18 x y z$ ✓. So

$\frac{1}{3} (3 x - 2 y + z) (9 x^{2} + 4 y^{2} + z^{2} + 6 x y + 2 y z - 3 z x)$

(x) The squares are $(2 x)^{2}, (3 y)^{2}, (6 z)^{2}$ , so a perfect square would have to be $(2 x + 3 y + 6 z)^{2} = 4 x^{2} + 9 y^{2} + 36 z^{2} + 12 x y + 36 y z + 24 z x$ . As printed, the coefficients of $x y$ and $z x$ are swapped (12xz and 24xy), and then the expression is not a perfect square. With the intended terms:

$4 x^{2} + 9 y^{2} + 36 z^{2} + 12 x y + 36 y z + 24 z x = (2 x + 3 y + 6 z)^{2}$

(xi) Rearranged: $27 u^{3} - \frac{9 u ^{2}}{2} + \frac{u}{4} - \frac{1}{216}$ . With $a = 3 u$ , $b = \frac{1}{6}$ : $3 a^{2} b = \frac{27 u ^{2}}{6} = \frac{9 u ^{2}}{2}$ ✓, $3 a b^{2} = \frac{9 u}{36} = \frac{u}{4}$ ✓. So it is $(3 u - \frac{1}{6})^{3}$ .

(i) $(2 y + \frac{1}{4 y})^{2}$ (ii) $(3 m - \frac{1}{5 n}) (3 m + \frac{1}{5 n})$ (iii) $(3 b - \frac{1}{4 b}) (9 b^{2} + \frac{3}{4} + \frac{1}{16 b ^{2}})$ (iv) $(x + \frac{1}{2}) (x + \frac{1}{3})$ (v) $(3 u - \frac{1}{5})^{3}$ (vi) $(4 y + \frac{z}{5}) (16 y^{2} - \frac{4 y z}{5} + \frac{z ^{2}}{25})$ (vii) $(p + 3 q + r) (p^{2} + 9 q^{2} + r^{2} - 3 pq - 3 q r - r p)$ (viii) $(3 m - 2)^{2}$ (ix) $\frac{1}{3} (3 x - 2 y + z) (9 x^{2} + 4 y^{2} + z^{2} + 6 x y + 2 y z - 3 z x)$ (x) $(2 x + 3 y + 6 z)^{2}$ (with 12xy + 24zx) (xi) $(3 u - \frac{1}{6})^{3}$

Simplify the following: (i)

\frac{4 x ^{2} + 4 x + 1}{4 x ^{2} - 1}

(ii)

\frac{9 ( 3 a ^{3} - 24 b ^{3} )}{9 a ^{2} - 36 b ^{2}}

(iii)

\frac{s ^{3} + 125 t ^{3}}{s ^{2} - 2 s t - 35 t ^{2}}

(Assume that the denominators are not equal to 0.)

Solution

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

(ii) Numerator: $9 \times 3 (a^{3} - 8 b^{3}) = 27 (a - 2 b) (a^{2} + 2 ab + 4 b^{2})$ . Denominator: $9 (a^{2} - 4 b^{2}) = 9 (a - 2 b) (a + 2 b)$ .

$\frac{27 ( a - 2 b ) ( a ^{2} + 2 ab + 4 b ^{2} )}{9 ( a - 2 b ) ( a + 2 b )} = \frac{3 ( a ^{2} + 2 ab + 4 b ^{2} )}{a + 2 b}$

(iii) Numerator: $(s + 5 t) (s^{2} - 5 s t + 25 t^{2})$ . Denominator: $(s - 7 t) (s + 5 t)$ .

$\frac{s ^{2} - 5 s t + 25 t ^{2}}{s - 7 t}$

(i) $\frac{2 x + 1}{2 x - 1}$ (ii) $\frac{3 ( a ^{2} + 2 ab + 4 b ^{2} )}{a + 2 b}$ (iii) $\frac{s ^{2} - 5 s t + 25 t ^{2}}{s - 7 t}$

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)

25 a^{2} - 30 ab + 9 b^{2}

(ii)

36 s^{2} - 49 t^{2}

Solution

Area = length × breadth, so we factor the area.

(i) $25 a^{2} - 30 ab + 9 b^{2} = (5 a - 3 b)^{2} = (5 a - 3 b) (5 a - 3 b)$ . Length and breadth are both $(5 a - 3 b)$ : the rectangle is a square.

(ii) $36 s^{2} - 49 t^{2} = (6 s + 7 t) (6 s - 7 t)$ . Length $= 6 s + 7 t$ , breadth $= 6 s - 7 t$ .

(i) Length = breadth = 5a − 3b (a square). (ii) Length = 6s + 7t, breadth = 6s − 7t.

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)

6 a^{2} - 24 b^{2}

(ii)

3 p s^{2} - 15 p s + 12 p

Solution

Volume = length × breadth × height, so we split the volume into three factors.

(i) $6 a^{2} - 24 b^{2} = 6 (a^{2} - 4 b^{2}) = 6 (a + 2 b) (a - 2 b)$ . Possible dimensions: 6, $(a + 2 b)$ and $(a - 2 b)$ (or, e.g., 2, $3 (a + 2 b)$ and $(a - 2 b)$ ).

(ii) $3 p s^{2} - 15 p s + 12 p = 3 p (s^{2} - 5 s + 4) = 3 p (s - 1) (s - 4)$ . Possible dimensions: $3 p$ , $(s - 1)$ and $(s - 4)$ .

(i) 6, (a + 2b), (a − 2b) (ii) 3p, (s − 1), (s − 4)

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 runs all round the outside of the playground. Then the outer boundary is a square of side $40 + 2 s$ metres (s on each side).

The hatched band is the path of width s around the 40 m square

$Area of path = (40 + 2 s)^{2} - 4 0^{2} = (1600 + 160 s + 4 s^{2}) - 1600 = 4 s^{2} + 160 s$

Or, using $a^{2} - b^{2} = (a + b) (a - b)$ : $(40 + 2 s + 40) (40 + 2 s - 40) = (80 + 2 s) (2 s) = 4 s^{2} + 160 s$ ✓

(If the path is laid inside the playground along its edge instead, the area is $4 0^{2} - (40 - 2 s)^{2} = 160 s - 4 s^{2}$ .)

Area of the path = $4 s^{2} + 160 s = 4 s (s + 40)$ square metres.

If a number plus its reciprocal equals

\frac{10}{3}

, find the number.

Solution

Let the number be $x$ ( $x \neq = 0$ ).

$x + \frac{1}{x} = \frac{10}{3} \Rightarrow 3 x^{2} + 3 = 10 x \Rightarrow 3 x^{2} - 10 x + 3 = 0$

Split $- 10 x$ as $- 9 x - x$ (since $(- 9) (- 1) = 9 = 3 \times 3$ ):

$3 x^{2} - 9 x - x + 3 = 3 x (x - 3) - 1 (x - 3) = (3 x - 1) (x - 3) = 0$

So $x = 3$ or $x = \frac{1}{3}$ . Check: $3 + \frac{1}{3} = \frac{10}{3}$ ✓

The number is 3 or $\frac{1}{3}$ .

A rectangular pool has area

2 x^{2} + 7 x + 3

square hastas. If its width is

2 x + 1

hastas, find its length. Hasta was a unit used to measure length.

Solution

Factor the area: $2 \times 3 = 6$ , and $7 = 6 + 1$ , so

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

$Length = \frac{Area}{Width} = \frac{( 2 x + 1 ) ( x + 3 )}{2 x + 1} = x + 3$

The length is (x + 3) hastas.

If both

x - 2

and

x - \frac{1}{2}

are factors of

p x^{2} + 5 x + r

, show that p = r.

Solution

If $x - 2$ is a factor, the expression is 0 at $x = 2$ ; if $x - \frac{1}{2}$ is a factor, it is 0 at $x = \frac{1}{2}$ .

$4 p + 10 + r = 0 (1)$

$\frac{p}{4} + \frac{5}{2} + r = 0 \Rightarrow p + 10 + 4 r = 0 (2)$

Subtracting (2) from (1): $3 p - 3 r = 0$ , so $p = r$ .

(In fact both equal −2: then $- 2 x^{2} + 5 x - 2 = - (2 x - 1) (x - 2)$ ✓.)

Subtracting the two conditions gives 3p − 3r = 0, so p = r (= −2).

If a + b + c = 5 and ab + bc + ca = 10, then prove that

a^{3} + b^{3} + c^{3} - 3 ab c = - 25

Solution

First find $a^{2} + b^{2} + c^{2}$ :

$a^{2} + b^{2} + c^{2} = (a + b + c)^{2} - 2 (ab + b c + c a) = 25 - 20 = 5$

Now use identity (8):

$a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) [(a^{2} + b^{2} + c^{2}) - (ab + b c + c a)] = 5 \times (5 - 10) = - 25$

$a^{3} + b^{3} + c^{3} - 3 ab c = 5 \times (5 - 10) = - 25$ , as required.

By factoring the expression, check that

n^{3} - n

is always divisible by 6 for all natural numbers n. Give reasons.

Solution

$n^{3} - n = n (n^{2} - 1) = (n - 1) n (n + 1)$

This is the product of three consecutive integers.

Among any two consecutive integers one is even, so the product is divisible by 2.
Among any three consecutive integers one is a multiple of 3, so the product is divisible by 3.

Since 2 and 3 have no common factor, the product is divisible by $2 \times 3 = 6$ . Examples: $n = 2$ : $1 \times 2 \times 3 = 6$ ; $n = 5$ : $4 \times 5 \times 6 = 120 = 6 \times 20$ . (For n = 1 it is 0, which is divisible by 6.)

$n^{3} - n = (n - 1) n (n + 1)$ is a product of three consecutive integers, which always contains a multiple of 2 and a multiple of 3, so it is divisible by 6.

Find the value of (i)

x^{3} + y^{3} - 12 x y + 64

, when x + y = −4 (ii)

x^{3} - 8 y^{3} - 36 x y - 216

, when x = 2y + 6

Solution

(i) Notice $64 = 4^{3}$ and $12 x y = 3 \cdot x \cdot y \cdot 4$ . So with $a = x$ , $b = y$ , $c = 4$ :

$x^{3} + y^{3} + 4^{3} - 3 (x) (y) (4) = (x + y + 4) (x^{2} + y^{2} + 16 - x y - 4 y - 4 x)$

Since $x + y = - 4$ , the first factor $x + y + 4 = 0$ . So the value is 0.

(ii) $x = 2 y + 6$ means $x - 2 y - 6 = 0$ . Take $a = x$ , $b = - 2 y$ , $c = - 6$ : $a^{3} + b^{3} + c^{3} = x^{3} - 8 y^{3} - 216$ and $3 ab c = 3 (x) (- 2 y) (- 6) = 36 x y$ . So the expression is $a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) (\dots) = (x - 2 y - 6) (\dots) = 0$ .

(i) 0 (ii) 0

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