Find the Exact Value (2^3)^-2

${(2^{3})}^{- 2}$

Multiply the exponents in ${(2^{3})}^{- 2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$2^{3 \cdot - 2}$

Multiply $3$ by $- 2$ .

$2^{- 6}$

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{2^{6}}$

Raise $2$ to the power of $6$ .

$\frac{1}{64}$

The result can be shown in multiple forms.

Exact Form:

$\frac{1}{64}$

Decimal Form:

$0.015625$