Let A = {(27)^{-3}⁄9^{-3}}^{1⁄3}
We know a^{m}⁄b^{m} = (a⁄b)^{m}
∴ A = {(27⁄9)^{-3}}^{1⁄3}
We know
(a^{m})^{n} = a^{mn}
∴ A = (27⁄9)^{-3 x 1⁄3}
= (27⁄9)^{-1} = (3)^{-1}
= 1⁄3. Ans.

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Let A = (√32 - √5)^{1⁄3}.(√32 + √5)^{1⁄3}
We Know a^{m} x b^{m} = (ab)^{m}
∴ A = {(√32 - √5)(√32 + √5)}^{1⁄3}
We Know
(a - b)(a + b) = a^{2} - b^{2}
∴ A = {(√32)^{2} - (√5)^{2}}^{1⁄3}
We Know
(√a)^{2} = a;
∴ A = {32 - 5}^{1⁄3} = (27)^{1⁄3}
We know 27 = 3 x 3 x 3 = 3^{3}
∴ A = (3^{3})^{1⁄3} = (3^{3 x 1⁄3}) [since (a^{m})^{n} = a^{mn}]
= 3^{1} = 3. Ans.

Solved Example 4 of Negative Exponents

Evaluate : 9^{5⁄2} - 3.4^{0} - (1⁄81)^{-1⁄2}

Solution to Example 4 of Negative Exponents:

Let A = 9^{5⁄2} - 3.(4)^{0} - (1⁄81)^{-1⁄2}
We know
9 = 3 x 3 = 3^{2}; (4)^{0} = 1; (1⁄81) = {1⁄(9 x 9)} = {1⁄(9)^{2} } = (9)^{-2};
∴ A = (3^{2})^{5⁄2} - 3.(1) - (9^{-2})^{-1⁄2}
= 3^{2 x 5⁄2} - 3 - (9^{-2 x -1⁄2}) [since (a^{m})^{n} = a^{mn}]
= 3^{5} - 3 - (9^{1}) = 3 x 3 x 3 x 3 x 3 - 3 - 9 = 243 - 12 = 231. Ans.

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If (9^{n} x 3^{2} x 3^{n} - 27^{n})⁄(3^{3m} x 2^{3}) = 3^{-3}, Prove that (m - n) = 1.

For Answers, see at the bottom of the page.

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