Exercise 3.2: Algebraic Expressions (Cubes and Expressions Involving Cubes):
1. Find the cube with the help of formulae:
(1)$2x^2 + 3y^2$
Solution: $(2x^2 + 3y^2)^3$
$= (2x^2)^3 + 3(2x^2)^2(3y^2) + 3(2x^2)(3y^2)^2 + (3y^2)^3$
$= 8x^6 + 3(4x^4)(3y^2) + 3(2x^2)(9y^4) + 27y^6$
$= 8x^6 + 36x^4y^2 + 54x^2y^4 + 27y^6$
Answer:$8x^6 + 36x^4y^2 + 54x^2y^4 + 27y^6$
(2)$7m^2 – 2n$
Solution: $(7m^2 – 2n)^3$
$= (7m^2)^3 – 3(7m^2)^2(2n) + 3(7m^2)(2n)^2 – (2n)^3$
$= 343m^6 – 3(49m^4)(2n) + 3(7m^2)(4n^2) – 8n^3$
$= 343m^6 – 294m^4n + 84m^2n^2 – 8n^3$
Answer:$343m^6 – 294m^4n + 84m^2n^2 – 8n^3$
(3)$2a – b – 3c$
Solution: $(2a – b – 3c)^3$
$= \{(2a – b) – 3c\}^3$
$= (2a – b)^3 – 3(2a – b)^2(3c) + 3(2a – b)(3c)^2 – (3c)^3$
$= \{(2a)^3 – 3(2a)^2(b) + 3(2a)(b)^2 – b^3\} – 9c(4a^2 – 4ab + b^2) + 3(2a – b)(9c^2) – 27c^3$
$= (8a^3 – 12a^2b + 6ab^2 – b^3) – (36a^2c – 36abc + 9b^2c) + (54ac^2 – 27bc^2) – 27c^3$
$= 8a^3 – b^3 – 27c^3 – 12a^2b + 6ab^2 – 36a^2c – 9b^2c + 54ac^2 – 27bc^2 + 36abc$
Answer:$8a^3 – b^3 – 27c^3 – 12a^2b + 6ab^2 – 36a^2c – 9b^2c + 54ac^2 – 27bc^2 + 36abc$
2. Simplify:
(1)$(7x + 3b)^3 – (5x + 3b)^3 – 6x(7x + 3b)(5x + 3b)$
Solution: Let $7x + 3b = m$ and $5x + 3b = n$.
Subtracting, we get:
$m – n = (7x + 3b) – (5x + 3b) = 2x$
$\implies 6x = 3(2x) = 3(m – n)$
Given expression $= m^3 – n^3 – 3(m – n)mn$
$= m^3 – n^3 – 3mn(m – n)$
$= (m – n)^3$
$= (2x)^3$
$= 8x^3$
Answer:$8x^3$
(2)$(a + b + c)^3 – (a – b – c)^3 – 6(b + c)\{a^2 – (b + c)^2\}$
Solution: Let $a + b + c = x$ and $a – b – c = y$.
Subtracting, we get:
$x – y = (a + b + c) – (a – b – c) = 2(b + c)$
Multiplying, we get:
$xy = (a + (b + c))(a – (b + c)) = a^2 – (b + c)^2$
Now, $6(b + c) = 3 \cdot 2(b + c) = 3(x – y)$
Given expression $= x^3 – y^3 – 3(x – y)xy$
$= x^3 – y^3 – 3xy(x – y)$
$= (x – y)^3$
$= \{2(b + c)\}^3$
$= 8(b + c)^3$
Answer:$8(b + c)^3$
(3)$(m + n)^6 – (m – n)^6 – 12mn(m^2 – n^2)^2$
Solution: Let $(m + n)^2 = a$ and $(m – n)^2 = b$.
Subtracting, we get:
$a – b = (m + n)^2 – (m – n)^2 = 4mn$
Multiplying, we get:
$ab = (m + n)^2(m – n)^2 = \{(m + n)(m – n)\}^2 = (m^2 – n^2)^2$
Now, $12mn = 3 \cdot (4mn) = 3(a – b)$
Given expression $= a^3 – b^3 – 3(a – b)ab$
$= a^3 – b^3 – 3ab(a – b)$
$= (a – b)^3$
$= (4mn)^3$
$= 64m^3n^3$
Answer:$64m^3n^3$
(4) $(x + y)(x^2 – xy + y^2) + (y + z)(y^2 – yz + z^2) + (z + x)(z^2 – zx + x^2)$
Solution:
We know that $a^3 + b^3 = (a + b)(a^2 – ab + b^2)$.
Therefore, given expression $= (x^3 + y^3) + (y^3 + z^3) + (z^3 + x^3)$
$= 2x^3 + 2y^3 + 2z^3$
$= 2(x^3 + y^3 + z^3)$
Answer:$2(x^3 + y^3 + z^3)$
(5)$(2x + 3y – 4z)^3 + (2x – 3y + 4z)^3 + 12x\{4x^2 – (3y – 4z)^2\}$
Solution: Let $2x + 3y – 4z = p$ and $2x – 3y + 4z = q$.
Adding, we get:
$p + q = 4x$
Multiplying, we get:
$pq = \{2x + (3y – 4z)\}\{2x – (3y – 4z)\} = (2x)^2 – (3y – 4z)^2 = 4x^2 – (3y – 4z)^2$
Now, $12x = 3 \cdot (4x) = 3(p + q)$
Given expression $= p^3 + q^3 + 3(p + q)pq$
$= p^3 + q^3 + 3pq(p + q)$
$= (p + q)^3$
$= (4x)^3$
$= 64x^3$
Answer:$64x^3$
3. If $a – b = 5$ and $ab = 36$, what is the value of $a^3 – b^3$?
Solution:
We know that $a^3 – b^3 = (a – b)^3 + 3ab(a – b)$
$= (5)^3 + 3(36)(5)$
$= 125 + 540$
$= 665$
Answer:$665$
4. If $a^3 – b^3 = 513$ and $a – b = 3$, what is the value of $ab$?
Solution:
We know that $a^3 – b^3 = (a – b)^3 + 3ab(a – b)$
$\implies 513 = (3)^3 + 3(ab)(3)$
$\implies 513 = 27 + 9ab$
$\implies 9ab = 513 – 27$
$\implies 9ab = 486$
$\implies ab = \frac{486}{9} = 54$
Answer:$54$
5. If $x = 19$ and $y = -12$, find the value of $8x^3 + 36x^2y + 54xy^2 + 27y^3$.
Solution:
Given, $x = 19$ and $y = -12$.
Given expression $= 8x^3 + 36x^2y + 54xy^2 + 27y^3$
$= (2x)^3 + 3(2x)^2(3y) + 3(2x)(3y)^2 + (3y)^3$
$= (2x + 3y)^3$
$= \{2(19) + 3(-12)\}^3$ [Substituting values]
$= (38 – 36)^3$
$= (2)^3$
$= 8$
Answer:$8$
6. If $a = 15$, what is the value of $8a^3 + 60a^2 + 150a + 130$?
Solution:
Given, $a = 15$.
Given expression $= 8a^3 + 60a^2 + 150a + 130$
$= (2a)^3 + 3(2a)^2(5) + 3(2a)(5)^2 + 5^3 + 5$
$= (2a + 5)^3 + 5$
$= \{2(15) + 5\}^3 + 5$ [Substituting $a = 15$]
$= (30 + 5)^3 + 5$
$= (35)^3 + 5$
$= 42875 + 5 = 42880$
Answer:$42880$
7. If $a + b = m$, $a^2 + b^2 = n$, and $a^3 + b^3 = p^3$, show that $m^3 + 2p^3 = 3mn$.
Solution:
Given, $a + b = m$, $a^2 + b^2 = n$, and $a^3 + b^3 = p^3$.
L.H.S. $= m^3 + 2p^3$
$= (a + b)^3 + 2(a^3 + b^3)$
$= (a^3 + 3a^2b + 3ab^2 + b^3) + 2a^3 + 2b^3$
$= 3a^3 + 3a^2b + 3ab^2 + 3b^3$
$= 3(a^3 + a^2b + ab^2 + b^3)$
$= 3\{a^2(a + b) + b^2(a + b)\}$
$= 3(a + b)(a^2 + b^2)$
$= 3mn = \text{R.H.S.}$ (Shown)
8. If $a + b = 3$ and $ab = 2$, find the values of (a) $a^2 – ab + b^2$ and (b) $a^3 + b^3$.
Solution:
(a) Given expression $= a^2 – ab + b^2$
$= (a + b)^2 – 2ab – ab$
$= (a + b)^2 – 3ab$
$= (3)^2 – 3(2)$
$= 9 – 6$
$= 3$
(b) $a^3 + b^3 = (a + b)(a^2 – ab + b^2)$
$= 3 \times 3$
$= 9$
Answer: (a) $3$, (b) $9$
9. If $a – b = 5$ and $ab = 36$, find the values of (a) $a^2 + ab + b^2$ and (b) $a^3 – b^3$.
Solution:
(a) Given expression $= a^2 + ab + b^2$
$= (a – b)^2 + 2ab + ab$
$= (a – b)^2 + 3ab$
$= (5)^2 + 3(36)$
$= 25 + 108$
$= 133$
(b) $a^3 – b^3 = (a – b)(a^2 + ab + b^2)$
$= 5 \times 133 = 665$
Answer: (a) $133$, (b) $665$
10. If $m + \frac{1}{m} = a$, find the value of $m^3 + \frac{1}{m^3}$.
Solution:
Given, $m + \frac{1}{m} = a$.
Given expression $= m^3 + \frac{1}{m^3}$
$= \left(m + \frac{1}{m}\right)^3 – 3 \cdot m \cdot \frac{1}{m} \left(m + \frac{1}{m}\right)$
$= a^3 – 3a$
Answer:$a^3 – 3a$
11. If $x – \frac{1}{x} = p$, find the value of $x^3 – \frac{1}{x^3}$.
Solution:
Given, $x – \frac{1}{x} = p$.
Given expression $= x^3 – \frac{1}{x^3}$
$= \left(x – \frac{1}{x}\right)^3 + 3 \cdot x \cdot \frac{1}{x} \left(x – \frac{1}{x}\right)$
$= p^3 + 3p$
Answer:$p^3 + 3p$
12. If $a – \frac{1}{a} = 1$, show that $a^3 – \frac{1}{a^3} = 4$.
Solution:
Given, $a – \frac{1}{a} = 1$.
L.H.S. $= a^3 – \frac{1}{a^3}$
$= \left(a – \frac{1}{a}\right)^3 + 3 \cdot a \cdot \frac{1}{a} \left(a – \frac{1}{a}\right)$
$= (1)^3 + 3(1)$
$= 1 + 3 = 4 = \text{R.H.S.}$ (Shown)
13. If $a + b + c = 0$, show that,
(a)$a^3 + b^3 + c^3 = 3abc$
Solution:
Given, $a + b + c = 0$
$\implies a + b = -c$
$\implies (a + b)^3 = (-c)^3$ [Cubing both sides]
$\implies a^3 + b^3 + 3ab(a + b) = -c^3$
$\implies a^3 + b^3 + 3ab(-c) = -c^3$
$\implies a^3 + b^3 – 3abc = -c^3$
$\implies a^3 + b^3 + c^3 = 3abc$ (Shown)
(b)$\frac{(b + c)^2}{3bc} + \frac{(c + a)^2}{3ca} + \frac{(a + b)^2}{3ab} = 1$
Solution:
Given, $a + b + c = 0$
Therefore, $b + c = -a$, $c + a = -b$, $a + b = -c$.
From (a), we have $a^3 + b^3 + c^3 = 3abc$.
L.H.S. $= \frac{(-a)^2}{3bc} + \frac{(-b)^2}{3ca} + \frac{(-c)^2}{3ab}$
$= \frac{a^2}{3bc} + \frac{b^2}{3ca} + \frac{c^2}{3ab}$
$= \frac{a^3 + b^3 + c^3}{3abc}$
$= \frac{3abc}{3abc} = 1 = \text{R.H.S.}$ (Shown)
14. If $p – q = r$, show that $p^3 – q^3 – r^3 = 3pqr$.
Solution:
Given, $p – q = r$
$\implies (p – q)^3 = r^3$ [Cubing both sides]
$\implies p^3 – q^3 – 3pq(p – q) = r^3$
$\implies p^3 – q^3 – 3pqr = r^3$
$\implies p^3 – q^3 – r^3 = 3pqr$ (Shown)
15. If $2x – \frac{2}{x} = 3$, show that $8\left(x^3 – \frac{1}{x^3}\right) = 63$.
Solution:
Given, $2\left(x – \frac{1}{x}\right) = 3$
$\implies x – \frac{1}{x} = \frac{3}{2}$
L.H.S. $= 8\left(x^3 – \frac{1}{x^3}\right)$
$= 8 \left\{ \left(x – \frac{1}{x}\right)^3 + 3 \cdot x \cdot \frac{1}{x} \left(x – \frac{1}{x}\right) \right\}$
$= 8 \left\{ \left(\frac{3}{2}\right)^3 + 3 \left(\frac{3}{2}\right) \right\}$
$= 8 \left\{ \frac{27}{8} + \frac{9}{2} \right\}$
$= 8 \times \frac{27 + 36}{8}$
$= 8 \times \frac{63}{8}$
$= 63 = \text{R.H.S.}$ (Shown)
16. If $a = \sqrt{6} + \sqrt{5}$, find the value of $\frac{a^6 – 1}{a^3}$.
Solution:
Given, $a = \sqrt{6} + \sqrt{5}$
$\implies \frac{1}{a} = \frac{1}{\sqrt{6} + \sqrt{5}}$
$= \frac{\sqrt{6} – \sqrt{5}}{(\sqrt{6} + \sqrt{5})(\sqrt{6} – \sqrt{5})}$
$= \frac{\sqrt{6} – \sqrt{5}}{6 – 5}$
$= \sqrt{6} – \sqrt{5}$
Subtracting, we get:
$a – \frac{1}{a} = (\sqrt{6} + \sqrt{5}) – (\sqrt{6} – \sqrt{5}) = 2\sqrt{5}$
Given expression $= \frac{a^6 – 1}{a^3} = \frac{a^6}{a^3} – \frac{1}{a^3} = a^3 – \frac{1}{a^3}$
$= \left(a – \frac{1}{a}\right)^3 + 3 \cdot a \cdot \frac{1}{a} \left(a – \frac{1}{a}\right)$
$= (2\sqrt{5})^3 + 3(2\sqrt{5})$
$= 8 \times 5\sqrt{5} + 6\sqrt{5}$
$= 40\sqrt{5} + 6\sqrt{5}$
$= 46\sqrt{5}$
Answer:$46\sqrt{5}$
17. $x – \frac{1}{x} = \sqrt{3}$ where $x \neq 0$
(1) Prove that, $x^2 – \sqrt{3}x = 1$
Solution:
Given, $x – \frac{1}{x} = \sqrt{3}$
$\implies \frac{x^2 – 1}{x} = \sqrt{3}$
$\implies x^2 – 1 = \sqrt{3}x$
$\implies x^2 – \sqrt{3}x = 1$ (Proved)
(2) Prove that $23\left(x^2 + \frac{1}{x^2}\right) = 5\left(x^4 + \frac{1}{x^4}\right)$
Solution:
Given, $x – \frac{1}{x} = \sqrt{3}$
• $x^2 + \frac{1}{x^2} = \left(x – \frac{1}{x}\right)^2 + 2 = (\sqrt{3})^2 + 2 = 3 + 2 = 5$
• $x^4 + \frac{1}{x^4} = \left(x^2 + \frac{1}{x^2}\right)^2 – 2 = (5)^2 – 2 = 25 – 2 = 23$
Now,
L.H.S. $= 23\left(x^2 + \frac{1}{x^2}\right) = 23 \times 5 = 115$
R.H.S. $= 5\left(x^4 + \frac{1}{x^4}\right) = 5 \times 23 = 115$
Therefore, L.H.S. $=$ R.H.S. (Proved)
(3) Find the value of $x^6 + \frac{1}{x^6}$
Solution:
From (b), we have $x^2 + \frac{1}{x^2} = 5$.
Given expression $= x^6 + \frac{1}{x^6} = (x^2)^3 + \left(\frac{1}{x^2}\right)^3$
$= \left(x^2 + \frac{1}{x^2}\right)^3 – 3 \cdot x^2 \cdot \frac{1}{x^2} \left(x^2 + \frac{1}{x^2}\right)$
$= (5)^3 – 3(5) = 125 – 15 = 110$
Answer:$110$