Answer
$t=8$ or $t=-27$
Work Step by Step
$ t^{\frac{2}{3}}+t^{\frac{1}{3}}-6=0\qquad$...substitute $t^{\frac{1}{3}}$ for $u$ so that $u^{2}=t^{\frac{2}{3}}$
$ u^{2}+u-6=0\qquad$... solve with the Quadratic formula.
$u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$u=\displaystyle \frac{-1\pm\sqrt{1+24}}{2}$
$u=\displaystyle \frac{-1\pm\sqrt{25}}{2}$
$u=\displaystyle \frac{-1\pm 5}{2}$
$u=\displaystyle \frac{-1+5}{2}=\frac{4}{2}=2$ or $u=\displaystyle \frac{-1-5}{2}=\frac{-6}{2}=-3$
Bring back $t^{\frac{1}{3}}=u$.
$t^{\frac{1}{3}}=2$ or $ t^{\frac{1}{3}}=-3\qquad$...raise both sides of both expressions to the third power
$(t^{\frac{1}{3}})^{3}=2^{3}$ or $(t^{\frac{1}{3}})^{3}=(-3)^{3} $
$t=8$ or $t=-27$