which equation shows a valid step in solving $sqrt3{2x - 6}+sqrt3{2x + 6}=0$?\n$left(sqrt3{2x…

which equation shows a valid step in solving $sqrt3{2x - 6}+sqrt3{2x + 6}=0$?\n$left(sqrt3{2x - 6}\right)^2=left(sqrt3{2x + 6}\right)^2$\n$left(sqrt3{2x - 6}\right)^2=left(-sqrt3{2x + 6}\right)^2$\n$left(sqrt3{2x - 6}\right)^3=left(sqrt3{2x + 6}\right)^3$\n$left(sqrt3{2x - 6}\right)^3=left(-sqrt3{2x + 6}\right)^3$

which equation shows a valid step in solving $sqrt3{2x - 6}+sqrt3{2x + 6}=0$?\n$left(sqrt3{2x - 6}\right)^2=left(sqrt3{2x + 6}\right)^2$\n$left(sqrt3{2x - 6}\right)^2=left(-sqrt3{2x + 6}\right)^2$\n$left(sqrt3{2x - 6}\right)^3=left(sqrt3{2x + 6}\right)^3$\n$left(sqrt3{2x - 6}\right)^3=left(-sqrt3{2x + 6}\right)^3$

Answer

Explanation:

Step1: Isolate one of the cube - root terms

Given $\sqrt[3]{2x - 6}+\sqrt[3]{2x + 6}=0$, we can isolate $\sqrt[3]{2x - 6}$ as $\sqrt[3]{2x - 6}=-\sqrt[3]{2x + 6}$.

Step2: Cube both sides

Cubing both sides of the equation $\sqrt[3]{2x - 6}=-\sqrt[3]{2x + 6}$, we get $(\sqrt[3]{2x - 6})^3=(-\sqrt[3]{2x + 6})^3$.

Answer:

$(\sqrt[3]{2x - 6})^3=(-\sqrt[3]{2x + 6})^3$