Example

Solve for $w$: $\quad 2w^{2/3}=3w^{1/3}+20$

Solution

The substitution $u=w^{1/3}$ transforms the equation to quadratic:

The new equation can be solved by factoring:

The quadratic equation has two solutions for the variable $u$, but we are interested in the solutions for the variable $w$. To find those, we must resubstitute and resolve for each $u$-solution.

These solutions should be checked in the original equation. They both check out! The original equation has two solutions: $w=-125/8$ or $w=4$.