Simplify (4^-1+2^-3)^-2

${(4^{- 1} + 2^{- 3})}^{- 2}$

Simplify each term.

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

${(\frac{1}{4} + 2^{- 3})}^{- 2}$

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

${(\frac{1}{4} + \frac{1}{2^{3}})}^{- 2}$

Raise $2$ to the power of $3$ .

${(\frac{1}{4} + \frac{1}{8})}^{- 2}$

To write $\frac{1}{4}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

${(\frac{1}{4} \cdot \frac{2}{2} + \frac{1}{8})}^{- 2}$

Write each expression with a common denominator of $8$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{1}{4}$ by $\frac{2}{2}$ .

${(\frac{2}{4 \cdot 2} + \frac{1}{8})}^{- 2}$

Multiply $4$ by $2$ .

${(\frac{2}{8} + \frac{1}{8})}^{- 2}$

Combine the numerators over the common denominator.

${(\frac{2 + 1}{8})}^{- 2}$

Add $2$ and $1$ .

${(\frac{3}{8})}^{- 2}$

Change the sign of the exponent by rewriting the base as its reciprocal.

${(\frac{8}{3})}^{2}$

Apply the product rule to $\frac{8}{3}$ .

$\frac{8^{2}}{3^{2}}$

Raise $8$ to the power of $2$ .

$\frac{64}{3^{2}}$

Raise $3$ to the power of $2$ .

$\frac{64}{9}$

The result can be shown in multiple forms.

Exact Form:

$\frac{64}{9}$

Decimal Form:

$7.\overline{1}$

Mixed Number Form:

$7 \frac{1}{9}$