Simplify 1/((2y^2z)^-3)

$\frac{1}{{(2 y^{2} z)}^{- 3}}$

Step 1

Move ${(2 y^{2} z)}^{- 3}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

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

Step 2

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

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

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

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

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

Step 3

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

$8 {(y^{2})}^{3} z^{3}$

Step 4

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

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

$8 y^{2 \cdot 3} z^{3}$

Multiply $2$ by $3$ .

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