Question 941183
*[tex \large \sqrt[3]{2x-4} - 2 = 0]


Cubing both sides of the equation does *not* give *[tex \large 2x-4 + 8 = 0] (or 2x-4 - 8 = 0, for that matter) -- look up binomial theorem. Cubing both sides instead gives you *[tex \large 2x-4 - 3 \cdot 2 \cdot \sqrt[3]{(2x-4)^2} + 3 \cdot 4 \cdot \sqrt[3]{2x-4} - 8 = 0], which does not look as easy to solve.


A simpler way to solve is to add 2 to both sides, giving *[tex \large \sqrt[3]{2x-4} = 2]. Then cube both sides to get *[tex \large 2x-4 = 8], or x = 6.