#### Answer

$t=1$

#### Work Step by Step

$ t^{\frac{1}{3}}+2t^{\frac{1}{6}}=3\qquad$... add $-3$ to both sides.
$ t^{\frac{1}{3}}+2t^{\frac{1}{6}}-3=0\qquad$...substitute $t^{\frac{1}{6}}$ for $u$ so that $u^{2}=t^{\frac{1}{3}}$
Note that $t^{\frac{1}{6}}$ is the sixth root of $t$, a positive number.
$ u^{2}+2u-3=0\qquad$... solve with the Quadratic formula.
$u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
$u=\displaystyle \frac{-2\pm\sqrt{4+12}}{2}$
$u=\displaystyle \frac{-2\pm\sqrt{16}}{2}$
$u=\displaystyle \frac{-2\pm 4}{2}$
$u=\displaystyle \frac{-2+4}{2}=\frac{2}{2}=1$ or $u=\displaystyle \frac{-2-4}{2}=\frac{-6}{2}=-3$
Bring back $t^{\frac{1}{6}}=u$.
$t^{\frac{1}{6}}=1$ or $t^{\frac{1}{3}}=-3$
We discard $-3$ because the sixth root of $t$ is nonnegative.
$t^{\frac{1}{6}}=1 $
$(t^{\frac{1}{6}})^{6}=1^{6}$
$t=1$