WBBSE Solutions For Class 8 Maths Algebra Chapter 3 Cubes

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Algebra Chapter 3 Cubes

⇒ If length of each side of a cube is x unit then its volume is x3 cubic units.

⇒ If length of each side of a cube is (a + b) unit then its volume is (a + b)3

⇒ (a + b)3 = (a + b) (a + b)2

⇒ (a + b)3 = (a + b) (a2 + 2ab + b2)

⇒ (a + b)3  = a(a2 + 2ab + b2) + b(a2 + 2ab + b2)

⇒ (a + b)3  = a3 + 2a2b + ab2 + a2b + 2ab2 + b3

∴ (a + b)3  = a3 + 3a2b + 3ab2 + b3

⇒ (a – b)3 = (a – b) (a – b)2

⇒ (a – b)3 = (a – b) (a2 – 2ab + b2)

⇒ (a – b)3 = a(a2 – 2ab + b2) – b(a2 – 2ab + b2)

⇒ (a – b)3= a3 – 2a2b + ab2 – a2b + 2ab2 – b3

⇒ (a – b)3= a3 – 3a2b + 3ab2 – b3

Necessary Formulae:

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  1. (a + b)3 = a3 + 3a2b + 3ab2 + b3 = a3 + b3 + 3ab (a + b) = a3 + b3 – 3ab (a – b)
  2. (a – b)3 = a3 – 3a2b + 3ab2 – b3
  3. a3+ b3 = (a + b)3 – 3ab (a + b) = (a + b) (a2 – ab + b2)
  4. a3 – b3 = (a – b)3 + 3ab (a – b) = (a – b) (a2 + ab + b2)
  5. a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c3 – ab- bc – ca) = \(\frac{1}{2}\)(a + b + c) {(a – b)2 + (b − c)2 + (c − a)2}

WBBSE Solutions For Class 8 Maths Algebra Chapter 3 Cubes

Algebra Chapter 3 Cubes Examples

Example 1. Find the value of

  1. (105)3
  2. (99)3

Solution:

1. (105)3 = (100+ 5)3

(a + b)3 = (100+ 5)3

Here a = 100, b= 5.

∴ (a + b)3 = a3 + 3a2b + 3ab2 + b3

= (100)3 + 3 x (100)2 x 5 + 3 x 100 x (5)2 + (5)3

= 1000000 + 150000+ 7500 +125

= 1157625

∴(105)3 = 1157625

The value of (105)3 = 1157625

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2. (99)3 = (100 – 1)3  = (a – b)3

Here a = 100, b= 1.

∴(a – b)3 = a3 – 3a2b + 3ab2 – b3

= (100)3 -3 x (100)2 x 1 + 3 x 100 x (1)2 – ( 1)3

= 1000000 – 30000 + 300 – 1

= 970299

∴ (99)3 = 970299

The value of (99)3 = 970299

Example 2. Find the cube of

  1. 3a + 5b
  2. 9x – 7y
  3. \(\frac{a}{2}\)–\(\frac{b}{3}\)
  4. a +2b – 3c

Solution:

1. (3a + 5b)3

∴ (a + b)3 = a3 + 3a2b + 3ab2 + b3

(3a + 5b)3 = (3a)3 + 3x (3a)2 x 5b + 3 x 3a x (5b)2 + (5b)3

(3a + 5b)3 = 27a3 + 135a2b + 150ab2 + 125b3

2. (9x – 7y)3

∴(a – b)3 = a3 – 3a2b + 3ab2 – b3

(9x – 7y)3 = (9x)3 – 3 x (9x)2 x 7y + 3 x 9x × (7y)2 – (7y)3

(9x – 7y)3 = 729x3 – 1701x2y + 1323xy3 – 343y3

3. \(\left(\frac{a}{2}-\frac{b}{3}\right)^3\)

= \(\left(\frac{a}{2}\right)^3-3 \times\left(\frac{a}{2}\right)^2 \times \frac{b}{3}+3 \times \frac{a}{2} \times\left(\frac{b}{3}\right)^2-\left(\frac{b}{3}\right)^3\)

= \(\frac{a^3}{8}-3 \times \frac{a^2}{4} \times \frac{b}{3}+3 \times \frac{a}{2} \times \frac{b^2}{9_3}-\frac{b^3}{27}=\frac{a^3}{8}-\frac{a^2 b}{4}+\frac{a b^2}{6}-\frac{b^3}{27}\)

4. (a + 2b – 3c)3

= {(a + 2b) – 3c)3

= (a + 2b)3 – 3 x (a + 2b)2 x 3c + 3 x (a + 2b) x (3c)2 – (3c)3

= a3 + 3 x a2 × 2b + 3 x a x (2b)2 + (2b)3 – 3 {a2 + 2.a.2b + (2b)2} x 3c +(3a + 6b) x 9c3 – 27c3

= a3 + 6a2b + 12ab2 + 8b3 – 9a2c – 36abc – 36b2c + 27ac2 + 54bc2 – 27c3

= a3 + 8b3 – 27c3 + 6a2b + 12ab2 – 9a2c – 36abc – 36b2c + 27ac2 + 54bc2 – 27c3

Example 3. If a – 2b = 5, then find the value of a3 – 8b3 – 30ab

Solution: a3 – 8b3 – 30ab

= (a)3 – (2b)3 – 30ab = (a – 2b)3 + 3 x a x 2b (a – 2b) – 30ab

= (5)3 + 6ab x 5 – 30ab [a – 2b = 5]

= 125 + 30ab – 30ab = 125

Alternating method: a3 – 8b3 – 30ab

= (a)3 – (2b)3 – 3 x a x 2b. (5)

=a3 – (2b)3 – 3 x a x 2b x (a – 2b)

= (a – 2b)3 = (5)3 = 125

The value of a3 – 8b3 – 30ab = 125

Example 4. Simplify: 7.4 x 7.4 x 7.4 + 3 x 7.4 x 7.4 x 2.6 + 3 x 7.4 x 2.6 x 2.6 + 2.6 x 2.6 x 2.6

Solution: 7.4 x 7.4 x 7.4 + 3 x 74 x 74 x 2.6 + 3 x 74 x 26 x 2.6+ 2.6 x 2.6 x 2.6

= (7.4)3 + 3 x (7.4)2 x 2.6 + 3 x 74 x (2.6)2 + (2.6)2 = (7.4 + 2.6)3 = (10)3 = 1000

7.4 x 7.4 x 7.4 + 3 x 7.4 x 7.4 x 2.6 + 3 x 7.4 x 2.6 x 2.6 + 2.6 x 2.6 x 2.6 = 1000

Example 5. If a3 + b3 + c3 = 3abc and a≠b≠c, then find the value of (a + b + c)

Solution: a3 + b3 + c3 = 3abc

⇒ a3 + b3 + c3 – 3abc = 0

⇒ (a + b + c)(a2 + b2 + c2 – ab- bc – ca) = 0

either, a + b + c = 0 or, a2 + b2 + c2 – ab – bc- ca = 0

⇒ 2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca = 0

⇒ (a2 – 2ab + b2) + (b2 – 2bc + c2) + (c2 – 2ca + a2) = 0

⇒ (a – b)2 + (b – c)2 + (c – a)2 = 0

If sum of two or more square are equal to zero then each square will be zero.

(a – b)2 = 0
(a – b)2 = 0
⇒ a – b = 0
⇒ a = b

(b – c)2 = 0
(b – c)2 = 0
⇒ b – c = 0
⇒ b = c

(c – a)2 = 0
⇒ (c – a)2 = 0

∴ a = b = c

But according to condition a ≠ b ≠ c

i.e. a2 + b2 + c2 – ab- bc – ca ≠ 0

∴ a + b + c = 0

The value of (a + b + c) = 0

Example 6. If 2a + \(\frac{1}{3a}\) = 4, then find the value of 27a3 + \(\frac{1}{8 a^3}\)

Solution: \(2 a+\frac{1}{3 a}=4\Rightarrow \frac{3}{2}\left(2 a+\frac{1}{3 a}\right)=4 \times \frac{3}{2} \Rightarrow 3 a+\frac{1}{2 a}=6\)

\(27 a^3+\frac{1}{8 a^3}=(3 a)^3+\left(\frac{1}{2 a}\right)^3=\left(3 a+\frac{1}{2 a}\right)^3-3 \times 3 a \times \frac{1}{2 a}\left(3 a+\frac{1}{2 a}\right)\)

= \((6)^3-\frac{9}{2} \times 6^3=216-27=189\)

Example 7. If m + n = 5 and mn = 6, then prove that (m2 + n2) (m3 + n3) = 455

Solution: (m2 + n2) (m3 + n3) = {(m + n)2 – 2mn} {(m + n)3 – 3mn (m + n)}

= {(5)2 – 2 x 6} {(5)3 – 3 x 6 x 5}= (25 – 12) (125 – 90)= 13 x 35 = 455

Example 8. Find the value of 64a3 – 144a2 + 108a – 50 if a = \(\frac{3}{2}\)

Solution:

let f(a)= 64a3 – 144a2 + 108a – 50

f(a) = (4a)3 – 3 x (4a)2 x 3 + 3 x 4a x (3)2 – (3)2 + (3)3 – 50

f(a)= (4a – 3)3 + 27 – 50  [a = \(\frac{3}{2}\)]

f(\(\frac{3}{2}\)) = \(\left(4^2 \times \frac{3}{2}-3\right)^3-23\)

f(\(\frac{3}{2}\)) = (6-3)3 – 23 = (3)3 – 23 = 27 – 23 = 4

The value of 64a3 – 144a2 + 108a – 50 = 4

Example 9. Simplify:

1. (4a2 – 25b2)(4a2 + 10ab + 25b2)(4a2 – 10ab + 25b2)

Solution: (4a2 – 25b2) (4a2 + 10ab + 25b2) (4a2 – 10ab + 25b2)

= {(2a)2 – (5b)2} {(2a)2 + 2a.5b + (5b)2} {(2a)2 – 2a.5b + (5b)2}

= (2a +5b) (2a – 5b) {(2a)2 + 2a.5b + (5b)2) {(2a)2 – 2a.5b + (5b)2}

= (2a +5b) {(2a)2 – 2a.5b + (5b)2} (2a – 5b) {(2a)2 + 2a.5b + (5b)2}

= {(2a)3 + (5b)3} {(2a)3 – (5b)3}

= (8a3 + 125b3)(8a3 – 125b3) Here (a+b × a-b) = a2-b2

= (8a3)2 -(125b3)2 = 64a6 – 15625b6.

(4a2 – 25b2)(4a2 + 10ab + 25b2)(4a2 – 10ab + 25b2) = (8a3)2 -(125b3)2 = 64a6 – 15625b6.

2. (x + 3) (x2 – 3x + 9)- (2x – 5) (4x2 + 10x + 25) – (x – 4) (x2 + 4x + 16)

Solution: (x + 3)(x2 – 3x+9)- (2x-5) (4x2+10x + 25) – (x – 4) (x2 + 4x + 16)

= (x + 3) (x2 – x.3 + 32) (2x – 5) {(2x)2 + 2x.5+ (5)2} – (x − 4) {x2 + x.4 + (4)2}

= (x3 + 33) – {(2x)3 – (5)3} – {x3 – 43}

= x3 + 27 – 8x3 + 125 – x3 + 64

= 216 – 8x3

(x + 3) (x2 – 3x + 9)- (2x – 5) (4x2 + 10x + 25) – (x – 4) (x2 + 4x + 16) = 216 – 8x3

Example 10. If \(\frac{a}{b}+{b}{a}\) = 1, then find the value of (a3 – b3)

Solution: \(\frac{a}{b}+{b}{a}\) = -1

⇒ \(\frac{a^2+b^2}{a b}=-1\)

⇒ a2 + b2 = -ab

⇒  a2 + ab + b2 = 0

a3 – b3 = (a – b) (a2 + ab + b2) = (a – b) (0) = 0

The value of (a3 – b3) = 0

Example 11. Resolve into factors:

1. a3 – 9b3 – 3ab(a – b)

Solution: a3 – 9b3 – 3ab (a – b) = a3 – b3 – 3ab(a – b) – 8b3

= (a – b)3 – (2b)3

= (a – b-2b) {(a – b)2 + (a – b).2b + (2b)2}

= (a – 3b) (a2 – 2ab + b2 + 2ab – 2b2 + 4b2)

= (a – 3b) (a2 + 3b2)

a3 – 9b3 – 3ab(a – b) = (a – 3b) (a2 + 3b2)

2. a12 – b2=12

Solution: \(\left(a^6\right)^2-\left(b^6\right)^2=\left(a^6+b^6\right)\left(a^6-b^6\right)\)

= \(\left\{\left(a^2\right)^3+\left(b^2\right)^3\right\}\left\{\left(a^3\right)^2-\left(b^3\right)^2\right\}\)

= \(\left(a^2+b^2\right)\left\{\left(a^2\right)^2-a^2 b^2+\left(b^2\right)^2\right\}\left(a^3+b^3\right)\left(a^3-b^3\right)\)

= \(\left(a^2+b^2\right)\left(a^4-a^2 b^2+b^4\right)(a+b)\left(a^2-a b+b^2\right)(a-b)\left(a^2+a b+b^2\right)\)

= \((a+b)(a-b)\left(a^2+b^2\right)\left(a^2+a b+b^2\right)\left(a^2-a b+b^2\right)\left(a^4-a^2 b^2+b^4\right)\)

3. \(\frac{a^3}{64}-\frac{64}{a^3}\)

Solution: = \(\left(\frac{a}{4}\right)^3-\left(\frac{4}{a}\right)^3=\left(\frac{a}{4}-\frac{4}{a}\right)\left\{\left(\frac{a}{4}\right)^2+\frac{a}{4} \cdot \frac{4}{a}+\left(\frac{4}{a}\right)^2\right\}\)

= \(\left(\frac{a}{4}-\frac{4}{a}\right)\left\{\left(\frac{a}{4}\right)^2+2 \cdot \frac{a}{4} \cdot \frac{b}{4}+\left(\frac{4}{a}\right)^2-1\right\}=\left(\frac{a}{4}-\frac{4}{a}\right)\left\{\left(\frac{a}{4}+\frac{4}{a}\right)^2-(1)^2\right\}\)

= \(\left(\frac{a}{4}-\frac{4}{a}\right)\left(\frac{a}{4}+\frac{4}{a}+1\right)\left(\frac{a}{4}+\frac{4}{a}-1\right)\)

4. \(x^3-3 x^2 y+3 x y^2-28 y^3\)

Solution: \(\left(x^3-3 x^2 y+3 x y^2-y^3\right)-27 y^3\)

∴(a – b)3 = a3 – 3a2b + 3ab2 – b3

= \((x-y)^3-(3 y)^3=(x-y-3 y)\left\{(x-y)^2+(x-y) \cdot 3 y+(3 y)^2\right\}\)

= \((x-4 y)\left\{x^2-2 x y+y^2+3 x y-3 y^2+9 y^2\right\}\)

= \((x-4 y)\left(x^2+x y+7 y^2\right)\)

Example 12. Simplify: \((a-c)\left\{(a+b)^2+(a+b)(b+c)+(b+c)^2\right\}\)

Solution: \(\{(a+b)-(b+c)\}\left\{(a+b)^2+(a+b)(b+c)+(b+c)^2\right.\)

= \((a+b)^3-(b+c)^3=a^3+3 a^2 b+3 a b^2+b^3-b^3-3 b^2 c-3 b c^2-c^3\)

= \(a^3+3 a^2 b+3 a b^2-3 b^2 c-3 b c^2-c^3\)

\((a-c)\left\{(a+b)^2+(a+b)(b+c)+(b+c)^2\right\}\) = \(a^3+3 a^2 b+3 a b^2-3 b^2 c-3 b c^2-c^3\)

Example 13. Simplify: If \(x^2+\frac{1}{16 x^2}=4 \frac{1}{2}\), then \(8 x^3-\frac{1}{8 x^3}=\)?

Solution: \(x^2+\frac{1}{16 x^2}=4 \frac{1}{2} \Rightarrow(x)^2+\left(\frac{1}{4 x}\right)^2=\frac{9}{2} \Rightarrow\left(x-\frac{1}{4 x}\right)^2+2 \cdot x \cdot \frac{1}{4 x}=\frac{9}{2}\)

⇒ \(\left(x-\frac{1}{4 x}\right)^2=\frac{9}{2}-\frac{1}{2}=4 \Rightarrow x-\frac{1}{4 x}=\sqrt{4}=2\)

⇒ \(2\left(x-\frac{1}{4 x}\right)=2 \times 2 \Rightarrow 2 x-\frac{1}{2 x}=4\)

\(8 x^3-\frac{1}{8 x^3}=(2 x)^3-\left(\frac{1}{2 x}\right)^3=\left(2 x-\frac{1}{2 x}\right)^3+3 \cdot 2 x \cdot \frac{1}{2 x}\left(2 x-\frac{1}{2 x}\right)\)

= (4)3 + 3.4 = 64 + 12 = 76

Example 14. If \(a+\frac{1}{a}=\sqrt{3}, \text { then } a^3+\frac{1}{a^3}=\text { ? }\)

Solution: \(a+\frac{1}{a}=\sqrt{3}\)

\(a^3+\frac{1}{a^3}=\left(a+\frac{1}{a}\right)^3-3 \cdot d \cdot \frac{1}{a}\left(a+\frac{1}{a}\right)=(\sqrt{3})^3-3 \sqrt{3}=3 \sqrt{3}-3 \sqrt{ } 3=0\)

Example 15. If a + b + c = 0, then \(\frac{a^2}{b c}+\frac{b^2}{c a}+\frac{c^2}{a b}\) =?

Solution: \(\frac{a^2}{b c}+\frac{b^2}{c a}+\frac{c^2}{a b}=\frac{a^3+b^3+c^3}{a b c}=\frac{\left(a^3+b^3+c^3-3 a b c\right)+3 a b c}{a b c}\)

= \(\frac{(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)+3 a b c}{a b c}\)

= \(\frac{(0)\left(a^2+b^2+c^2-a b-b c-c a\right)+3 a b c}{a b c}=\frac{0+3 a b c}{a b c}=3\)

Example 16. Choose the correct answer:

1. If a + b + c = 0, then a3 + b3 + c3 =?

  1. 3abc
  2. abc
  3. c
  4. None of these

Solution: a3 + b3 + c3

= (a + b)3 – 3ab (a + b) + c3

= (-c)3 – 3ab (-c) + c3 [ a + b + c = 0]

= c3 + 3abc + c3 = 3abc

∴ So the correct answer is 1. 3abc

2. If a – b = 1 and a3 – b3 = 61, then the value of ab is

  1. 10
  2. 20
  3. 30
  4. None of these

Solution: a3 – b3 = 61

⇒ (a – b)3 + 3ab (a – b) = 61

⇒ (1)3 + 3ab (1) = 61

⇒ 3ab = 61 – 1 = 60

⇒ ab = \(\frac{60}{3}\) = 20

∴ So the correct answer is 2. 20

3. If \(x^2+\frac{1}{16 x^2}=3 \frac{1}{2}\) then tha value of \(8 x^3+\frac{1}{8 x^3}\) is

  1. 66
  2. 56
  3. 76
  4. 52

Solution: \(x^2+\frac{1}{16 x^2}=\frac{7}{2}\)

⇒ \((x)^2+\left(\frac{1}{4 x}\right)^2=\frac{7}{2} \Rightarrow\left(x+\frac{1}{4 x}\right)^2-2 \cdot x \cdot \frac{1}{24 x}=\frac{7}{2} \)

⇒ \(\left(x+\frac{1}{4 x}\right)^2=\frac{7}{2}+\frac{1}{2}=4 \Rightarrow x+\frac{1}{4 x}=\sqrt{4}=2\)

⇒ \(2\left(x+\frac{1}{4 x}\right)=2 \times 2 \Rightarrow 2 x+\frac{1}{2 x}=4\)

\(8 x^3+\frac{1}{8 x^3}=(2 x)^3+\left(\frac{1}{2 x}\right)^3\)

= \(\left(2 x+\frac{1}{2 x}\right)^3-3 \cdot 2 x \cdot \frac{1}{2 x}\left(2 x+\frac{1}{2 x}\right)\)

= \((4)^3-3 \cdot 4=64-12=52\)

∴ The correct answer is 4. 52

Example 17. Write ‘True’ or ‘False’:

1. 1729 is a Hardy Ramanujam number.

Solution: The statement is true.

2. If \(3 x+\frac{3}{x}=2\), then the value od \(x^3+\frac{1}{x^3}=\frac{27}{89}\)

Solution: \(3 x+\frac{3}{x}=2 \Rightarrow 3\left(x+\frac{1}{x}\right)=2 \Rightarrow x+\frac{1}{x}=\frac{2}{3}\)

\(x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3 \cdot+\cdot \frac{1}{x}\left(x+\frac{1}{x}\right)=\left(\frac{2}{3}\right)^3-3 \cdot \frac{2}{3}\)

= \(\frac{8}{27}-2=\frac{8-54}{27}=-\frac{46}{27}\)

∴ Statement is false

3. \(p^3 q^3+1=(p q-1)\left(p^2 q^2+p q+1\right)\)

Solution: \(p^3 q^3+1=(p q)^3+(1)^3=(p q+1)\left\{(p q)^2-p q \cdot 1+(1)^2\right\}\)

= \((p q+1)\left(p^2 q^2-p q+1\right)\)

∴ The statement is false.

Example 18. Fill in the blanks:

1. If a + \(\frac{9}{a}\) = 3, then a3 + 27 = _________

Solution: a + \(\frac{9}{a}\) = 3

⇒ a2 + 9 = 3a

⇒ a2 – 3a + 9 = 0

⇒ a3 + 27 = (a)3 + (3)3

⇒ a3 + 27 = (a + 3) (a2 – a·3 + 32)

⇒ a3 + 27 = (a + 3) (a2 – 3a + 9)= (a + 3) (0) = 0

2. If t2 – 4t + 1 = 0, then t3 + \(\frac{1}{t^3}\) = ______

Solution: \(t^2-4 t+1=0 \Rightarrow t^2+1=4 t\)

⇒ \(\frac{t^2+1}{t}=4 \Rightarrow t+\frac{1}{t}=4\)

⇒ \(t^3+\frac{1}{t^3}=\left(t+\frac{1}{t}\right)^3-3 \cdot t \cdot \frac{1}{t}\left(t+\frac{1}{t}\right)=(4)^3-3 \cdot 4=52\)

3. (a + b)3 = a3 + b3 + _________

Solution: 3ab (a + b)

(a + b)3 = a3 + b3  = 3ab (a + b)

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