Question

simplify. assume all variables are positive.
(27yz)^{-\frac{2}{3}}
write your answer in the form a or \frac{a}{b}, where a and b are constants or variable expressions that have no variables in common. all exponents in your answer should be positive.

Explanation:

Step1: Apply power - of - product rule

$$(27yz)^{-\frac{2}{3}}=27^{-\frac{2}{3}}y^{-\frac{2}{3}}z^{-\frac{2}{3}}$$

Step2: Rewrite negative exponents as positive

$$27^{-\frac{2}{3}}y^{-\frac{2}{3}}z^{-\frac{2}{3}}=\frac{1}{27^{\frac{2}{3}}y^{\frac{2}{3}}z^{\frac{2}{3}}}$$

Step3: Evaluate $27^{\frac{2}{3}}$

First, find $27^{\frac{1}{3}}$. Since $3^3 = 27$, then $27^{\frac{1}{3}}=3$. So, $27^{\frac{2}{3}}=(27^{\frac{1}{3}})^2 = 3^2=9$.

Step4: Get the final form

$$\frac{1}{27^{\frac{2}{3}}y^{\frac{2}{3}}z^{\frac{2}{3}}}=\frac{1}{9y^{\frac{2}{3}}z^{\frac{2}{3}}}$$

Answer:

$\frac{1}{9y^{\frac{2}{3}}z^{\frac{2}{3}}}$