Answer
$x_{1}=27$ and $x_{2}=8$
Work Step by Step
Given $x^{2/3}-5x^{1/3}+6=0 \longrightarrow (x^{1/3})^2-5x^{1/3}+6=0$
Substituting $x^{1/3}=u$, we have: $(x^{1/3})^2-5x^{1/3}+6=0 \longrightarrow u^2-5u+6=0$
$a=1,\ b=-5,\ c=6$
Using the quadratic formula: $\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$, we have:
$\dfrac{-(-5)\pm \sqrt{(-5)^2-4\times 1\times 6}}{2\times 1} = \dfrac{5 \pm \sqrt{25-24}}{2} = \dfrac{5\pm \sqrt{1}}{2} = \dfrac{5\pm 1}{2}$
Therefore, we have that: $u_{1}=\dfrac{5 + 1}{2} = \dfrac{6}{2} = 3$ and $u_{2}=\dfrac{5-1}{2}=\dfrac{4}{2}=2$
Substituting: $x^{1/3}_{1}=u_{1}=3 \longrightarrow x_{1}=3^3=27$ and $x^{1/3}_{2}=u_{2}=2 \longrightarrow x_{2}=2^3=8$
