How do I solve for $x$ when $x^{\frac23}-3x^{\frac13}-10=0$?

How do I solve for $x$ when $x^{\frac23}-3x^{\frac13}-10=0$?

I can't factor anything and there aren't any like terms to combine, the only thing I could think of doing is turning the exponents into radicals but I don't think that helps.

How do I solve for $x$ in a situation like this?

Hint. Consider the substitution $y = x^{1/3}.$ Then we have $$y^2 - 3y - 10 = 0.$$
• factoring this gives be $y=-2$ and $y=5$, then substituting back the $x^{\frac13}$ gives me $x=-8$ and $x=125$, thank you. – jotam5326 Nov 22 '16 at 1:48