Simplify ((2x^0y^-2)/z)^-3

${(\frac{2 x^{0} y^{- 2}}{z})}^{- 3}$

Step 1

Move $y^{- 2}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

${(\frac{2}{z y^{2}})}^{- 3}$

Step 2

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

${(\frac{z y^{2}}{2})}^{3}$

Step 3

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $\frac{z y^{2}}{2}$ .

$\frac{{(z y^{2})}^{3}}{2^{3}}$

Apply the product rule to $z y^{2}$ .

$\frac{z^{3} {(y^{2})}^{3}}{2^{3}}$

Step 4

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

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

$\frac{z^{3} y^{2 \cdot 3}}{2^{3}}$

Multiply $2$ by $3$ .

$\frac{z^{3} y^{6}}{2^{3}}$

Step 5

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

$\frac{z^{3} y^{6}}{8}$