Find the value ${[4^{- 1} + 3^{- 1} + 6^{- 2}]}^{- 1}$

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Solution

Compute the given expression
Given ${[4^{- 1} + 3^{- 1} + 6^{- 2}]}^{- 1}$ can be written as ${[\frac{1}{4} + \frac{1}{3} + \frac{1}{36}]}^{- 1}$
$∴ \frac{1}{4} + \frac{1}{3} + \frac{1}{36} = \frac{9 + 12 + 1}{36} = \frac{22}{36}$
${[\frac{1}{4} + \frac{1}{3} + \frac{1}{36}]}^{- 1} = {[\frac{22}{36}]}^{- 1} = \frac{36}{22} = \frac{18}{11} \{∵ a^{- 1} = \frac{1}{a}\}$
Hence, the value of ${[4^{- 1} + 3^{- 1} + 6^{- 2}]}^{- 1}$ is $\frac{18}{11}$

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