Indices [Exponents] — Exercise 7(A)

$\Big(-\dfrac{2a^2}{b^3}\Big)^4$ is equal to :

1. $-\dfrac{16a^8}{b^{12}}$

2. $\dfrac{16a^8}{b^{12}}$

3. $\dfrac{2a^8}{b^{12}}$

4. $-\dfrac{2a^8}{b^{12}}$

Answer

Simplifying the expression :

$\Rightarrow \Big(-\dfrac{2a^2}{b^3}\Big)^4 = \Big(-2 \times \dfrac{a^2}{b^3}\Big)^4 \\[1em] = (-2)^4 \times \Big(\dfrac{a^2}{b^3}\Big)^4 = 16 \times \dfrac{a^{2 \times 4}}{b^{3 \times 4}} \\[1em] = \dfrac{16a^8}{b^{12}}.$

Hence, Option 2 is the correct option.

$\sqrt[4]{28} ÷ \sqrt[4]{7}$ is equal to :

1. $\sqrt{2}$

2. $\sqrt{4}$

3. $\sqrt[4]{2}$

4. $\sqrt[3]{2}$

Answer

Simplifying the expression :

$\Rightarrow \sqrt[4]{28} ÷ \sqrt[4]{7} = \dfrac{\sqrt[4]{28}}{\sqrt[4]{7}} \\[1em] = \Big(\dfrac{28}{7}\Big)^{\dfrac{1}{4}} = (4)^{\dfrac{1}{4}} \\[1em] = (2)^{2 \times \dfrac{1}{4}} = (2)^{\dfrac{1}{2}} \\[1em] = \sqrt{2}.$

Hence, Option 1 is the correct option.

$\Big(\dfrac{25}{9}\Big)^{-\dfrac{3}{2}}$ is equal to :

1. $\Big(\dfrac{5}{3}\Big)^3$

2. $\Big(\dfrac{5}{3}\Big)^2$

3. $\dfrac{27}{125}$

4. $\Big(\dfrac{3}{5}\Big)^{-2}$

Answer

Simplifying the expression :

$\Rightarrow \Big(\dfrac{25}{9}\Big)^{-\dfrac{3}{2}} = \Big[\Big(\dfrac{5}{3}\Big)^2\Big]^{-\dfrac{3}{2}} \\[1em] = \Big(\dfrac{5}{3}\Big)^{-3} = \Big(\dfrac{3}{5}\Big)^3 \\[1em] = \dfrac{27}{125}.$

Hence, Option 3 is the correct option.

$2 ÷ (243)^{-\dfrac{1}{5}}$ is equal to :

1. 2 ÷ 3

2. $\dfrac{1}{6}$

3. 12

4. 6

Answer

Simplifying the expression :

$\Rightarrow 2 ÷ (243)^{-\dfrac{1}{5}} = 2 ÷ (3^5)^{-\dfrac{1}{5}} \\[1em] = 2 ÷ 3^{-1} = 2 ÷ \dfrac{1}{3} \\[1em] = 2 \times 3 \\[1em] = 6.$

Hence, Option 4 is the correct option.

$(0.01)^{-\dfrac{1}{2}}$ :

1. 10

2. 0.1

3. $(0.1)^{\frac{1}{2}}$

4. 0.0001

Answer

Simplifying the expression :

$\Rightarrow (0.01)^{-\dfrac{1}{2}} = \Big(\dfrac{1}{100}\Big)^{-\dfrac{1}{2}} \\[1em] = \Big(\dfrac{1}{10^2}\Big)^{-\dfrac{1}{2}} = (10^{-2})^{-\dfrac{1}{2}} \\[1em] = 10^1 = 10.$

Hence, Option 1 is the correct option.

$3 \times (32)^{\dfrac{2}{5}} \times 7^0$ is equal to :

1. 0

2. 12

3. 1

4. 9

Answer

Simplifying the expression :

$\Rightarrow 3 \times (32)^{\dfrac{2}{5}} \times 7^0 = 3 \times (2^5)^{\dfrac{2}{5}} \times 7^0 \\[1em] = 3 \times 2^2 \times 1 \\[1em] = 3 \times 4 \\[1em] = 12.$

Hence, Option 2 is the correct option.

Evaluate :

$3^3 \times (243)^{-\dfrac{2}{3}} \times (9)^{-\dfrac{1}{3}}$

Answer

Simplifying the expression :

$\Rightarrow 3^3 \times (3^5)^{-\dfrac{2}{3}} \times (3^2)^{-\dfrac{1}{3}} \\[1em] = 3^3 \times (3)^{-\dfrac{10}{3}} \times (3)^{-\dfrac{2}{3}} \\[1em] = (3)^{3 + \Big(-\dfrac{10}{3}\Big) + \Big(-\dfrac{2}{3}\Big)} \\[1em] = (3)^{\dfrac{9 - 10 - 2}{3}} = (3)^{\dfrac{-3}{3}} \\[1em] = 3^{-1} = \dfrac{1}{3}.$

Hence, $3^3 \times (243)^{-\dfrac{2}{3}} \times (9)^{-\dfrac{1}{3}} = \dfrac{1}{3}$.

Evaluate :

$5^{-4} \times (125)^{\dfrac{5}{3}} ÷ (25)^{-\dfrac{1}{2}}$

Answer

Simplifying the expression :

$\Rightarrow 5^{-4} \times (125)^{\dfrac{5}{3}} ÷ (25)^{-\dfrac{1}{2}} = 5^{-4} \times (5^3)^{\dfrac{5}{3}} ÷ (5^2)^{-\dfrac{1}{2}} \\[1em] = 5^{-4} \times 5^5 ÷ 5^{-1} = 5^{-4} \times 5^5 ÷ \dfrac{1}{5} \\[1em] = 5^{-4} \times 5^5 \times 5^1 \\[1em] = 5^{-4 + 5 + 1} = 5^2 \\[1em] = 25.$

Hence, $5^{-4} \times (125)^{\dfrac{5}{3}} ÷ (25)^{-\dfrac{1}{2}} = 25.$

Evaluate :

$\Big(\dfrac{27}{125}\Big)^{\dfrac{2}{3}} \times \Big(\dfrac{9}{25}\Big)^{-\dfrac{3}{2}}$

Answer

Simplifying the expression :

$\Rightarrow \Big(\dfrac{27}{125}\Big)^{\dfrac{2}{3}} \times \Big(\dfrac{9}{25}\Big)^{-\dfrac{3}{2}} = \Big[\Big(\dfrac{3}{5}\Big)^3\Big]^{\dfrac{2}{3}} \times \Big[\Big(\dfrac{3}{5}\Big)^2\Big]^{-\dfrac{3}{2}} \\[1em] = \Big(\dfrac{3}{5}\Big)^2 \times \Big(\dfrac{3}{5}\Big)^{-3} = \Big(\dfrac{3}{5}\Big)^{2 + (-3)} \\[1em] = \Big(\dfrac{3}{5}\Big)^{-1} = \dfrac{5}{3} = 1\dfrac{2}{3}.$

Hence, $\Big(\dfrac{27}{125}\Big)^{\dfrac{2}{3}} \times \Big(\dfrac{9}{25}\Big)^{-\dfrac{3}{2}} = 1\dfrac{2}{3}$.

Evaluate :

$7^0 \times (25)^{-\dfrac{3}{2}} - 5^{-3}$

Answer

Simplifying the expression :

$\Rightarrow 7^0 \times (25)^{-\dfrac{3}{2}} - 5^{-3} = 1 \times (5^2)^{-\dfrac{3}{2}} - 5^{-3} \\[1em] = 1 \times 5^{-3} - 5^{-3} \\[1em] = 5^{-3} - 5^{-3} \\[1em] = 0.$

Hence, $7^0 \times (25)^{-\dfrac{3}{2}} - 5^{-3} = 0.$

Evaluate :

$\Big(\dfrac{16}{81}\Big)^{-\dfrac{3}{4}} \times \Big(\dfrac{49}{9}\Big)^{\dfrac{3}{2}} ÷ \Big(\dfrac{343}{216}\Big)^{\dfrac{2}{3}}$

Answer

Simplifying the expression :

$\Rightarrow \Big(\dfrac{16}{81}\Big)^{-\dfrac{3}{4}} \times \Big(\dfrac{49}{9}\Big)^{\dfrac{3}{2}} ÷ \Big(\dfrac{343}{216}\Big)^{\dfrac{2}{3}} = \Big[\Big(\dfrac{2}{3}\Big)^4\Big]^{-\dfrac{3}{4}} \times \Big[\Big(\dfrac{7}{3}\Big)^2\Big]^{\dfrac{3}{2}} ÷ \Big[\Big(\dfrac{7}{6}\Big)^3\Big]^{\dfrac{2}{3}} \\[1em] = \Big(\dfrac{2}{3}\Big)^{4 \times -\dfrac{3}{4}} \times \Big(\dfrac{7}{3}\Big)^{2 \times \dfrac{3}{2}} ÷ \Big(\dfrac{7}{6}\Big)^{3 \times \dfrac{2}{3}} \\[1em] = \Big(\dfrac{2}{3}\Big)^{-3} \times \Big(\dfrac{7}{3}\Big)^3 ÷ \Big(\dfrac{7}{6}\Big)^2 \\[1em] = \Big(\dfrac{3}{2}\Big)^3 \times \Big(\dfrac{7}{3}\Big)^3 \times \Big(\dfrac{6}{7}\Big)^2 \\[1em] = \dfrac{3^3 \times 7^3 \times 6^2}{2^3 \times 3^3 \times 7^2} \\[1em] = \dfrac{7 \times 6^2}{2^3} \\[1em] = \dfrac{252}{8} \\[1em] = 31.5$

Hence, $\Big(\dfrac{16}{81}\Big)^{-\dfrac{3}{4}} \times \Big(\dfrac{49}{9}\Big)^{\dfrac{3}{2}} ÷ \Big(\dfrac{343}{216}\Big)^{\dfrac{2}{3}} = 31.5$

Simplify :

$(8x^3 ÷ 125y^3)^{\dfrac{2}{3}}$

Answer

Simplifying the expression :

$\Rightarrow \Big(\dfrac{8x^3}{125y^3}\Big)^{\dfrac{2}{3}} =\Big[\dfrac{(2x)^3}{(5y)^3}\Big]^{\dfrac{2}{3}} \\[1em] = \Big[\Big(\dfrac{2x}{5y}\Big)^3\Big]^{\dfrac{2}{3}} = \Big(\dfrac{2x}{5y}\Big)^2 \\[1em] = \dfrac{4x^2}{25y^2}.$

Hence, $(8x^3 ÷ 125y^3)^{\dfrac{2}{3}} = \dfrac{4x^2}{25y^2}.$

Simplify :

(a + b)^-1.(a^-1 + b^-1)

Answer

Simplifying the expression :

$\Rightarrow (a + b)^{-1}(a^{-1} + b^{-1}) = \dfrac{1}{(a + b)} \times \Big(\dfrac{1}{a} + \dfrac{1}{b}\Big) \\[1em] = \dfrac{1}{(a + b)} \times \Big(\dfrac{b + a}{ab}\Big) \\[1em] = \dfrac{1}{ab}.$

Hence, $(a + b)^{-1}(a^{-1} + b^{-1}) = \dfrac{1}{ab}$.

Simplify :

$\dfrac{5^{n + 3} - 6 \times 5^{n + 1}}{9 \times 5^n - 5^n \times 2^2}$

Answer

Simplifying the expression :

$\Rightarrow \dfrac{5^{n + 3} - 6 \times 5^{n + 1}}{9 \times 5^n - 5^n \times 2^2} = \dfrac{5^n.5^3 - 6 \times 5^n \times 5^1}{5^n(9 - 2^2)} \\[1em] = \dfrac{5^n(5^3 - 6 \times 5)}{5^n(9 - 4)} = \dfrac{125 - 30}{5} = \dfrac{95}{5} \\[1em] = 19.$

Hence, $\dfrac{5^{n + 3} - 6 \times 5^{n + 1}}{9 \times 5^n - 5^n \times 2^2} = 19.$

Simplify :

$(3x^2)^{-3} \times (x^9)^{\dfrac{2}{3}}$

Answer

Simplifying the expression :

$\Rightarrow (3x^2)^{-3} \times (x^9)^{\dfrac{2}{3}} = \Big(\dfrac{1}{3x^2}\Big)^3 \times x^{9 \times \dfrac{2}{3}} \\[1em] = \dfrac{1}{27x^6} \times x^6 \\[1em] = \dfrac{1}{27}.$

Hence, $(3x^2)^{-3} \times (x^9)^{\dfrac{2}{3}} = \dfrac{1}{27}$.

Evaluate :

$\sqrt{\dfrac{1}{4}} + (0.01)^{-\dfrac{1}{2}} - (27)^{\dfrac{2}{3}}$

Answer

Simplifying the expression :

$\Rightarrow \sqrt{\dfrac{1}{4}} + (0.01)^{-\dfrac{1}{2}} - (27)^{\dfrac{2}{3}} = \dfrac{1}{2} + \Big(\dfrac{1}{100}\Big)^{-\dfrac{1}{2}} - (3^3)^{\dfrac{2}{3}} \\[1em] = \dfrac{1}{2} + \Big(\dfrac{1}{10^2}\Big)^{-\dfrac{1}{2}} - 3^2 = \dfrac{1}{2} + (10^{-2})^{-\dfrac{1}{2}} - 3^{3 \times \dfrac{2}{3}} \\[1em] = \dfrac{1}{2} + 10^{-2 \times -\dfrac{1}{2}} - 3^2 \\[1em] = \dfrac{1}{2} + 10 - 9 \\[1em] = \dfrac{1}{2} + 1 \\[1em] = \dfrac{1 + 2}{2} \\[1em] = \dfrac{3}{2} \\[1em] = 1\dfrac{1}{2}.$

Hence, $\sqrt{\dfrac{1}{4}} + (0.01)^{-\dfrac{1}{2}} - (27)^{\dfrac{2}{3}} = 1\dfrac{1}{2}.$

Evaluate :

$\Big(\dfrac{27}{8}\Big)^{\dfrac{2}{3}} - \Big(\dfrac{1}{4}\Big)^{-2} + 5^0$

Answer

Simplifying the expression :

$\Rightarrow \Big(\dfrac{27}{8}\Big)^{\dfrac{2}{3}} - \Big(\dfrac{1}{4}\Big)^{-2} + 5^0 = \Big[\Big(\dfrac{3}{2}\Big)^3\Big]^{\dfrac{2}{3}} - \Big(\dfrac{1}{2^2}\Big)^{-2} + 5^0\\[1em] = \Big(\dfrac{3}{2}\Big)^{3 \times \dfrac{2}{3}} - (2^2)^2 + 1 \\[1em] = \Big(\dfrac{3}{2}\Big)^2 - 2^4 + 1 \\[1em] = \dfrac{9}{4} - 16 + 1 \\[1em] = \dfrac{9 - 64 + 4}{4} \\[1em] = -\dfrac{51}{4}.$

Hence, $\Big(\dfrac{27}{8}\Big)^{\dfrac{2}{3}} - \Big(\dfrac{1}{4}\Big)^{-2} + 5^0 = -\dfrac{51}{4}$.

Simplify the following and express with positive index :

$\Big(\dfrac{3^{-4}}{2^{-8}}\Big)^{\dfrac{1}{4}}$

Answer

Simplifying the expression :

$\Rightarrow \Big(\dfrac{3^{-4}}{2^{-8}}\Big)^{\dfrac{1}{4}} = \dfrac{3^{-4\times \dfrac{1}{4}}}{2^{-8 \times \dfrac{1}{4}}} \\[1em] = \dfrac{3^{-1}}{2^{-2}} = \dfrac{\dfrac{1}{3}}{\dfrac{1}{2^2}} \\[1em] = \dfrac{2^2}{3} = \dfrac{4}{3}.$

Hence, $\Big(\dfrac{3^{-4}}{2^{-8}}\Big)^{\dfrac{1}{4}} = \dfrac{4}{3}$.

Simplify the following and express with positive index :

$\Big(\dfrac{27^{-3}}{9^{-3}}\Big)^{\dfrac{1}{5}}$

Answer

Simplifying the expression :

$\Rightarrow \Big(\dfrac{27^{-3}}{9^{-3}}\Big)^{\dfrac{1}{5}} = \Big[\dfrac{(3^3)^{-3}}{(3^2)^{-3}}\Big]^{\dfrac{1}{5}}\\[1em] = \Big(\dfrac{3^{3 \times -3}}{3^{2 \times -3}}\Big)^{\dfrac{1}{5}} = \Big(\dfrac{3^{-9}}{3^{-6}}\Big)^{\dfrac{1}{5}} \\[1em] = (3^{-9 - (-6)})^{\dfrac{1}{5}} = (3^{-9 + 6})^{\dfrac{1}{5}} \\[1em] = (3^{-3})^{\dfrac{1}{5}} = (3)^{-\dfrac{3}{5}} \\[1em] = \Big(\dfrac{1}{3^3}\Big)^{\dfrac{1}{5}} = \dfrac{1}{3^{\dfrac{3}{5}}}.$

Hence, $\Big(\dfrac{27^{-3}}{9^{-3}}\Big)^{\dfrac{1}{5}} = \dfrac{1}{3^{\dfrac{3}{5}}}$.

Simplify the following and express with positive index :

$(32)^{-\dfrac{2}{5}} ÷ (125)^{-\dfrac{2}{3}}$

Answer

Simplifying the expression :

$\Rightarrow (32)^{-\dfrac{2}{5}} ÷ (125)^{-\dfrac{2}{3}} = (2^5)^{-\dfrac{2}{5}} ÷ (5^3)^{-\dfrac{2}{3}} \\[1em] = (2)^{5 \times -\dfrac{2}{5}} ÷ (5)^{3 \times -\dfrac{2}{3}} \\[1em] = (2)^{-2} ÷ 5^{-2} \\[1em] = \dfrac{1}{2^2} ÷ \dfrac{1}{5^2} \\[1em] = \dfrac{1}{2^2} \times 5^2 \\[1em] = \dfrac{1}{4} \times 25 \\[1em] = \dfrac{25}{4} = 6\dfrac{1}{4}.$

Hence, $(32)^{-\dfrac{2}{5}} ÷ (125)^{-\dfrac{2}{3}} = 6\dfrac{1}{4}$.

Simplify the following and express with positive index :

[1 - {1 - (1 - n)^-1}^-1]^-1

Answer

Simplifying the expression :

$\Rightarrow [1 - {1 - (1 - n)^{-1}}^{-1}]^{-1} = \Big[1 - \text{\textbraceleft}1 - \dfrac{1}{1 - n}\text{\textbraceright}^{-1}\Big]^{-1} \\[1em] = \Big[1 - \text{\textbraceleft}\dfrac{1 - n - 1}{1 - n}\text{\textbraceright}^{-1}\Big]^{-1} \\[1em] = \Big[1 - \text{\textbraceleft}\dfrac{-n}{1 - n}\text{\textbraceright}^{-1}\Big]^{-1} \\[1em] = \Big[1 - \text{\textbraceleft}-\dfrac{1 - n}{n}\text{\textbraceright}\Big]^{-1} \\[1em] = \Big[1 + \dfrac{1 - n}{n}\Big]^{-1} \\[1em] = \Big[\dfrac{n + 1 - n}{n}\Big]^{-1} \\[1em] = \Big[\dfrac{1}{n}\Big]^{-1} \\[1em] = n.$

Hence, $[1 - {1 - (1 - n)^{-1}}^{-1}]^{-1} = n$.

If 2160 = 2^a.3^b.5^c, find a, b and c. Hence, calculate the value of 3^a × 2^-b × 5^-c.

Answer

Factorizing 2160, we get :

⇒ 2160 = 2⁴ × 3³ × 5¹

⇒ 2^a.3^b.5^c = 2⁴ × 3³ × 5¹

⇒ a = 4, b = 3 and c = 1.

Substituting values of a, b and c in 3^a × 2^-b × 5^-c, we get :

$\Rightarrow 3^4 \times 2^{-3} \times 5^{-1} = 81 \times \dfrac{1}{2^3} \times \dfrac{1}{5} \\[1em] = 81 \times \dfrac{1}{8} \times \dfrac{1}{5} \\[1em] = \dfrac{81}{40} = 2\dfrac{1}{40}.$

Hence, a = 4, b = 3 and c = 1 and 3^a × 2^-b × 5^-c = $2\dfrac{1}{40}$.

If 1960 = 2^a.5^b.7^c, calculate the value of 2^-a.7^b.5^-c.

Answer

Factorizing 1960, we get :

⇒ 1960 = 2³.5¹.7²

⇒ 2^a.5^b.7^c = 2³.5¹.7²

⇒ a = 3, b = 1 and c = 2.

Substituting values of a, b and c in 2^-a.7^b.5^-c, we get :

$\Rightarrow 2^{-3} \times 7^1 \times 5^{-2} = \dfrac{1}{2^3} \times 7 \times \dfrac{1}{5^2} \\[1em] = \dfrac{1}{8} \times 7 \times \dfrac{1}{25} \\[1em] = \dfrac{7}{200}.$

Hence, 2^-a.7^b.5^-c = $\dfrac{7}{200}$.

Simplify :

$\dfrac{8^{3a} \times 2^5 \times 2^{2a}}{4 \times 2^{11a} \times 2^{-2a}}$

Answer

Simplifying the expression :

$\Rightarrow \dfrac{8^{3a} \times 2^5 \times 2^{2a}}{4 \times 2^{11a} \times 2^{-2a}} = \dfrac{(2^3)^{3a} \times 2^5 \times 2^{2a}}{2^2 \times 2^{11a} \times 2^{-2a}} \\[1em] = \dfrac{2^{9a} \times 2^5 \times 2^{2a}}{2^2 \times 2^{11a} \times 2^{-2a}} \\[1em] = \dfrac{2^{9a + 5 + 2a}}{2^{2 + 11a + (-2a)}} \\[1em] = \dfrac{2^{11a + 5}}{2^{9a + 2}} \\[1em] = 2^{(11a + 5) - (9a + 2)} \\[1em] = 2^{11a - 9a + 5 - 2} \\[1em] = 2^{2a + 3}.$

Hence, $\dfrac{8^{3a} \times 2^5 \times 2^{2a}}{4 \times 2^{11a} \times 2^{-2a}} = 2^{2a + 3}$.

Simplify :

$\dfrac{3 \times 27^{n + 1} + 9 \times 3^{3n - 1}}{8 \times 3^{3n} - 5 \times 27^n}$

Answer

Simplifying the expression

$\Rightarrow \dfrac{3 \times 27^{n + 1} + 9 \times 3^{3n - 1}}{8 \times 3^{3n} - 5 \times 27^n} = \dfrac{3 \times (3^3)^{n + 1} + (3^2) \times 3^{3n - 1}}{8 \times 3^{3n} - 5 \times (3^3)^n}\\[1em] = \dfrac{3 \times 3^{3(n + 1)} + 3^{2 + 3n - 1}}{8 \times 3^{3n} - 5 \times 3^{3n}} \\[1em] = \dfrac{3^{1 + 3(n + 1)} + 3^{3n + 1}}{3^{3n}(8 - 5)} \\[1em] = \dfrac{3^{1 + 3n + 3} + 3^{3n + 1}}{3 \times 3^{3n}} \\[1em] = \dfrac{3^{3n + 1}.3^3 + 3^{3n + 1}}{3^{3n + 1}} \\[1em] = \dfrac{3^{3n + 1}(3^3 + 1)}{3^{3n + 1}} \\[1em] = 3^3 + 1 \\[1em] = 27 + 1 \\[1em] = 28.$

Hence, $\dfrac{3 \times 27^{n + 1} + 9 \times 3^{3n - 1}}{8 \times 3^{3n} - 5 \times 27^n} = 28$.

Show that :

$\Big(\dfrac{a^m}{a^{-n}}\Big)^{m - n} \times \Big(\dfrac{a^n}{a^{-l}}\Big)^{n - l} \times (\dfrac{a^l}{a^{-m}}\Big)^{l - m}$ = 1

Answer

Solving L.H.S. of the above equation :

$\Rightarrow \Big(\dfrac{a^m}{a^{-n}}\Big)^{m - n} \times \Big(\dfrac{a^n}{a^{-l}}\Big)^{n - l} \times (\dfrac{a^l}{a^{-m}}\Big)^{l - m} \\[1em] \Rightarrow (a^{m - (-n)})^{m - n} \times (a^{n - (-l)})^{n - l} \times (a^{l - (-m)})^{l - m} \\[1em] \Rightarrow (a^{(m + n)})^{m - n} \times (a^{(n + l)})^{n - l} \times (a^{(l + m)})^{l - m} \\[1em] \Rightarrow a^{(m + n)(m - n)} \times a^{(n + l)(n - l)} \times a^{(l + m)(l - m)} \\[1em] \Rightarrow a^{m^2 - n^2} \times a^{n^2 - l^2} \times a^{l^2 - m^2} \\[1em] \Rightarrow a^{m^2 - n^2 + n^2 - l^2 +l^2 - m^2} \\[1em] \Rightarrow a^{m^2 - m^2 - n^2 + n^2 - l^2 + l^2} \\[1em] \Rightarrow a^0 \\[1em] \Rightarrow 1.$

Since, L.H.S. = R.H.S. = 1.

Hence, proved that $\Big(\dfrac{a^m}{a^{-n}}\Big)^{m - n} \times \Big(\dfrac{a^n}{a^{-l}}\Big)^{n - l} \times (\dfrac{a^l}{a^{-m}}\Big)^{l - m}$ = 1.

If a = x^{m + n}.y^l; b = x^{n + l}.y^m and c = x^{l + m}.yⁿ,

prove that : a^{m - n}.b^{n - l}.c^{l - m} = 1

Answer

Substituting values of a, b and c in L.H.S. of equation a^{m - n}.b^{n - l}.c^{l - m} = 1, we get :

⇒ a^{m - n}.b^{n - l}.c^{l - m} = (x^{m + n}.y^l)^{m - n}.(x^{n + l}.y^m)^{n - l}.(x^{l + m}.yⁿ)^{l- m}

= (x^{(m + n)(m - n)}.y^{l(m - n)}).(x^{(n + l)(n - l)}.y^{m(n - l)}).(x^{(l + m)(l - m)}.y^{n(l - m)})

= (x^{m2 - n2}).(x^{n2 - l2}).(x^{l2 - m2}).(y^{lm - ln}).(y^{mn - ml}).(y^{nl - nm})

= x^{m2 - n2 + n2 - l2 + l2 - m2}.y^{lm - ln + mn - ml + nl - nm}

= x⁰.y⁰

= 1.1

= 1.

Since, L.H.S. = R.H.S. = 1.

Hence, proved that a^{m - n}.b^{n - l}.c^{l- m} = 1.

Simplify :

$\Big(\dfrac{x^a}{x^b}\Big)^{a^2 + ab + b^2} \times \Big(\dfrac{x^b}{x^c}\Big)^{b^2 + bc + c^2} \times \Big(\dfrac{x^c}{x^a}\Big)^{c^2 + ca + a^2}$

Answer

Simplifying the expression :

$\Rightarrow \Big(\dfrac{x^a}{x^b}\Big)^{a^2 + ab + b^2} \times \Big(\dfrac{x^b}{x^c}\Big)^{b^2 + bc + c^2} \times \Big(\dfrac{x^c}{x^a}\Big)^{c^2 + ca + a^2} \\[1em] = (x^{a - b})^{a^2 + ab + b^2} \times (x^{b - c})^{b^2 + bc + c^2} \times (x^{c - a})^{c^2 + ca + a^2} \\[1em] = x^{(a - b)(a^2 + ab + b^2)} \times x^{(b - c)(b^2 + bc + c^2)} \times x^{(c - a)(c^2 + ca + a^2)} \\[1em] = x^{a^3 - b^3} \times x^{b^3 - c^3} \times x^{c^3 - a^3} \\[1em] = x^{a^3 - b^3 + b^3 - c^3 + c^3 - a^3} \\[1em] = x^0 \\[1em] = 1.$

Hence, $\Big(\dfrac{x^a}{x^b}\Big)^{a^2 + ab + b^2} \times \Big(\dfrac{x^b}{x^c}\Big)^{b^2 + bc + c^2} \times \Big(\dfrac{x^c}{x^a}\Big)^{c^2 + ca + a^2}$ = 1.

Simplify :

$\Big(\dfrac{x^a}{x^{-b}}\Big)^{a^2 - ab + b^2} \times \Big(\dfrac{x^b}{x^{-c}}\Big)^{b^2 - bc + c^2} \times \Big(\dfrac{x^c}{x^{-a}}\Big)^{c^2 - ca + a^2}$

Answer

Simplifying the expression :

$\Rightarrow \Big(\dfrac{x^a}{x^{-b}}\Big)^{a^2 - ab + b^2} \times \Big(\dfrac{x^b}{x^{-c}}\Big)^{b^2 - bc + c^2} \times \Big(\dfrac{x^c}{x^{-a}}\Big)^{c^2 - ca + a^2} \\[1em] = (x^{a - (-b)})^{a^2 - ab + b^2} \times (x^{b - (-c)})^{b^2 - bc + c^2} \times (x^{c - (-a)})^{c^2 - ca + a^2} \\[1em] = x^{(a + b)(a^2 - ab + b^2)} \times x^{(b + c)(b^2 - bc + c^2)} \times x^{(c + a)(c^2 - ca + a^2)} \\[1em] = x^{a^3 + b^3} \times x^{b^3 + c^3} \times x^{c^3 + a^3} \\[1em] = x^{a^3 + b^3 + b^3 + c^3 + c^3 + a^3} \\[1em] = x^{2a^3 + 2b^3 + 2c^3} \\[1em] = x^{2(a^3 + b^3 + c^3)}.$

Hence, $\Big(\dfrac{x^a}{x^{-b}}\Big)^{a^2 - ab + b^2} \times \Big(\dfrac{x^b}{x^{-c}}\Big)^{b^2 - bc + c^2} \times \Big(\dfrac{x^c}{x^{-a}}\Big)^{c^2 - ca + a^2} = x^{2(a^3 + b^3 + c^3)}.$