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in Exponents by (46.2k points)
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\(\frac{1}{(216)^{\frac{-2}{3}}}+\frac{1}{(256)^\frac{-3}{4}}+\frac{1}{(243)^{\frac{-1}{5}}}\) is equal to

(a) 103 

(b) 105 

(c) 107 

(d) 101

1 Answer

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Best answer

(a) 103

Given exp. = \(\frac{1}{{(6^3)}^{-\frac{2}{3}}}+\frac{1}{(4^4)^{-\frac{3}{4}}}+\frac{1}{(3^5)^{-\frac{1}{5}}}\)

\(\frac{1}{6^{-2}}+\frac{1}{4^{-3}}+\frac{1}{3^{-1}}\)

= 62 + 43 + 3 = 36 + 64 + 3 = 103

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