which equation shows a valid, practical step in solving $sqrt4{2x - 8}+sqrt4{2x + 8}=0$?\n$left(sqrt4{2x…

which equation shows a valid, practical step in solving $sqrt4{2x - 8}+sqrt4{2x + 8}=0$?\n$left(sqrt4{2x - 8}\right)^3=-left(sqrt4{2x + 8}\right)^3$\n$left(sqrt4{2x - 8}\right)^3=left(-sqrt4{2x + 8}\right)^3$\n$left(sqrt4{2x - 8}\right)^4=-left(sqrt4{2x + 8}\right)^4$\n$left(sqrt4{2x - 8}\right)^4=left(-sqrt4{2x + 8}\right)^4$

which equation shows a valid, practical step in solving $sqrt4{2x - 8}+sqrt4{2x + 8}=0$?\n$left(sqrt4{2x - 8}\right)^3=-left(sqrt4{2x + 8}\right)^3$\n$left(sqrt4{2x - 8}\right)^3=left(-sqrt4{2x + 8}\right)^3$\n$left(sqrt4{2x - 8}\right)^4=-left(sqrt4{2x + 8}\right)^4$\n$left(sqrt4{2x - 8}\right)^4=left(-sqrt4{2x + 8}\right)^4$

Answer

Explanation:

Step1: Isolate one of the radical terms

Given $\sqrt[4]{2x - 8}+\sqrt[4]{2x + 8}=0$, we can rewrite it as $\sqrt[4]{2x - 8}=-\sqrt[4]{2x + 8}$.

Step2: Eliminate the fourth - root

To get rid of the fourth - root, we raise both sides of the equation to the fourth power. According to the property $(a)^n=(b)^n$ when $a = b$, we have $(\sqrt[4]{2x - 8})^4=(-\sqrt[4]{2x + 8})^4$.

Answer:

$(\sqrt[4]{2x - 8})^4=(-\sqrt[4]{2x + 8})^4$