Solve ${[1^{2} + 2^{2} + 3^{2}]}^{- 3 / 2}$

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Solution

$\begin{matrix} {[1^{2} + 2^{2} + 3^{2}]}^{- 3 / 2} = {[1 + 4 + 9]}^{- 3 / 2} = 14^{- 3 / 2} = {(\frac{1}{14})}^{3 / 2} \end{matrix}$

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Solve [12+22+32]−3/2

[12+22+32]−3/2=[1+4+9]−3/2=14−3/2=(114)3/2

Solve ${[1^{2} + 2^{2} + 3^{2}]}^{- 3 / 2}$

$\begin{matrix} {[1^{2} + 2^{2} + 3^{2}]}^{- 3 / 2} = {[1 + 4 + 9]}^{- 3 / 2} = 14^{- 3 / 2} = {(\frac{1}{14})}^{3 / 2} \end{matrix}$