Answer
$-2(x-2)$ or, $4-2x$
Work Step by Step
Simplify $\sqrt[5] {(-32)(x-2)^5}$
For $n$ is even, we have $\sqrt [n] {a^n}=|a|$
For $n$ is odd, we have $\sqrt [n] {a^n}=a$
Thus, $\sqrt[5] {(-32)(x-2)^5}=\sqrt[5] {(-2)^5(x-2)^5}$
or, $=\sqrt[5] {(-2(x-2))^5}$
or, $=-2(x-2)$
or, $=-2x+4$
or, $=4-2x$
