[Math] Sum of cube roots of a quadratic

algebra-precalculuspolynomialsquadratics

If $a$ and $b$ are the roots of $x^2 -5x + 8 = 0$. How do I find $\sqrt[3]{a} + \sqrt[3]{b}$ without finding the roots?

I know how to evaluate $\sqrt[2]{a} + \sqrt[2]{b}$ by squaring and subbing for $a+b$ and $ab$ via sum and product of roots. But for this question, if I cube $\sqrt[3]{a} + \sqrt[3]{b}$ I'm left with radicals which are difficult to resolve, e.g. $\sqrt[3]{a^2b}$

How should I go about approaching this problem?

Edit: I've also tried letting $\sqrt[3]a+\sqrt[3]b=m$, which makes $a+b+3m\sqrt[3]{ab}=m^3$ (by rising everything to the power of $3$ and then substituting $\sqrt[3]a+\sqrt[3]b=m$ again), if that is of any help.

Best Answer

Hint:.

$$s^3=(\sqrt[3]a+\sqrt[3]b)^3=a+b+3\sqrt[3]{ab}(\sqrt[3]a+\sqrt[3]b)=5+6s$$