Answer
$x=\pm2\sqrt{2}$ and $x=\pm3\sqrt{3}$
Work Step by Step
$x^{4/3}-5x^{2/3}+6=0$
Let $x^{2/3}$ be equal to $u$:
$u=x^{2/3}$, so $u^{2}=x^{4/3}$:
Substitute $x^{4/3}$ by $u^{2}$ and $x^{2/3}$ by $u$ in the original equation:
$u^{2}-5u+6=0$
Solve this equation by factoring:
$(u-2)(u-3)=0$
Set both factors equal to $0$ and solve each individual equation for $u$:
$u-2=0$
$u=2$
$u-3=0$
$u=3$
Substitute $u$ back to $x^{2/3}$ and solve for $x$:
$u=2$
$x^{2/3}=2$
$x=2^{3/2}$
$x=\sqrt{2^{3}}=\pm2\sqrt{2}$
$u=3$
$x^{2/3}=3$
$x=3^{3/2}$
$x=\sqrt{3^{3}}=\pm3\sqrt{3}$
The solutions to this equation are $x=\pm2\sqrt{2}$ and $x=\pm3\sqrt{3}$
