$\displaystyle 64^{-\frac{2}{3}}$
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The key to this question is knowing/working out that $\displaystyle 64 = 4^3$
If you prefer you can use prime factors
$\displaystyle 64 = 2 \times 32$
$\displaystyle 32 = 2 \times 16$
$\displaystyle 16 = 2 \times 8$
$\displaystyle 8 = 2 \times 4$
$\displaystyle 4 = 2 \times 2$
Therefore $\displaystyle 64 = 2^6 = 2^{3+3} = 2^3 \cdot 2^3$
$\displaystyle (\sqrt[3]{64})^2 = [\sqrt[3]{2^3} \times \sqrt[3]{2^3}]^2$
Cubes and cube roots cancel to leave $\displaystyle (2 \times 2)^2 = 16
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