If $a = 1, b = 2$ then find the value of $(a^{b} + b^{a})^{- 1}$ .

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Solution

For a = 1 and b = 2,

$(a^{b} + b^{a})^{- 1} = (1^{2} + 2^{1})^{- 1} = (1 + 2)^{- 1} = 3^{- 1} = \frac{1}{3}$

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