Solve for M M=(-1/8)^(-3^-1)+(1/16)^(-4^-1)

$M = {(- \frac{1}{8})}^{- 3^{- 1}} + {(\frac{1}{16})}^{- 4^{- 1}}$

Simplify each term.

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

$M = {(- \frac{1}{8})}^{- \frac{1}{3}} + {(\frac{1}{16})}^{- 4^{- 1}}$

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

$M = {(- 8)}^{\frac{1}{3}} + {(\frac{1}{16})}^{- 4^{- 1}}$

Rewrite $- 8$ as ${(- 2)}^{3}$ .

$M = {({(- 2)}^{3})}^{\frac{1}{3}} + {(\frac{1}{16})}^{- 4^{- 1}}$

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

$M = {(- 2)}^{3 (\frac{1}{3})} + {(\frac{1}{16})}^{- 4^{- 1}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$M = {(- 2)}^{3 (\frac{1}{3})} + {(\frac{1}{16})}^{- 4^{- 1}}$

Rewrite the expression.

$M = {(- 2)}^{1} + {(\frac{1}{16})}^{- 4^{- 1}}$

Evaluate the exponent.

$M = - 2 + {(\frac{1}{16})}^{- 4^{- 1}}$

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

$M = - 2 + {(\frac{1}{16})}^{- \frac{1}{4}}$

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

$M = - 2 + 16^{\frac{1}{4}}$

Rewrite $16$ as $2^{4}$ .

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

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

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

$M = - 2 + 2^{1}$

Evaluate the exponent.

$M = - 2 + 2$

Add $- 2$ and $2$ .

$M = 0$