Answer
$w=64$ or $w=-8$
Work Step by Step
$ w^{\frac{2}{3}}-2w^{\frac{1}{3}}-8=0\qquad$...substitute $w^{\frac{1}{3}}$ for $u$ so that $u^{2}=w^{\frac{2}{3}}$
$ u^{2}-2u-8=0\qquad$... solve with the Quadratic formula.
$u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$u=\displaystyle \frac{2\pm\sqrt{4+32}}{2}$
$u=\displaystyle \frac{2\pm\sqrt{36}}{2}$
$u=\displaystyle \frac{2\pm 6}{2}$
$u=\displaystyle \frac{2+6}{2}=\frac{8}{2}=4$ or $u=\displaystyle \frac{2-6}{2}=\frac{-4}{2}=-2$
Bring back $w^{\frac{1}{3}}=u$.
$w^{\frac{1}{3}}=4$ or $ w^{\frac{1}{3}}=-2\qquad$...raise both sides of both expressions to the third power
$(w^{\frac{1}{3}})^{3}=4^{3}$ or $(w^{\frac{1}{3}})^{3}=(-2)^{3} $
$w=64$ or $w=-8$
