[tex](\sqrt[4]{2x-8})^4\text{= (-}\sqrt[4]{2x\text{ + 8}})^4\text{ (option D)}[/tex]
Explanation:[tex]\sqrt[4]{2x-8}\text{ + }\sqrt[4]{2x\text{ + 8}}\text{ = 0}[/tex][tex]\begin{gathered} \text{let 2x + 8 = y} \\ \sqrt[4]{y}\text{ + }\sqrt[4]{y}\text{ = 0} \\ \text{subtract }\sqrt[4]{y}\text{ from both sides}\colon \\ \sqrt[4]{y}\text{ +}\sqrt[4]{y}\text{ - }\sqrt[4]{y}\text{ =0- }\sqrt[4]{y}\text{ } \end{gathered}[/tex][tex]\begin{gathered} \sqrt[4]{y}\text{ + }0\text{= }0\text{ - }\sqrt[4]{y}\text{ } \\ \sqrt[4]{y}\text{ = - }\sqrt[4]{y}\text{ } \\ y\text{ = 2x + 8} \\ \sqrt[4]{2x-8}\text{ = - }\sqrt[4]{2x\text{ + 8}}\text{ } \end{gathered}[/tex][tex]\begin{gathered} the\text{ root on both side is fourth root} \\ To\text{ take it away on both sides, we would raise power to 4 on both sides} \\ (\sqrt[4]{2x-8})^4\text{= (-}\sqrt[4]{2x\text{ + 8}})^4\text{ (option D)} \end{gathered}[/tex]
