Sum of cube roots of a quadratic

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.

Solution 1:

Hint:.

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

Solution 2:

The solutions of the quadratic equation are not real numbers, so $\sqrt[3]{a}$ and $\sqrt[3]{b}$ are not canonically determined; therefore, the sum can take up to 9 different values.

Since $ab = 8$, once you choose one of the three possible values for $\sqrt[3]{a}$, you may want to consider only the value $\sqrt[3]{b} = 2/\sqrt[3]{a}$ for the second root. In that case, let $S = \sqrt[3]{a} + \sqrt[3]{b}$. Then $S^3 = a+b + 3\sqrt[3]{ab} S$, so $S$ is a solution of the cubic equation $S^3 = 6S+5 \Longleftrightarrow (S+1)(S^2-S-5) = 0$. The solutions are $S_1=-1$, $S_2=\frac{1}{2}(1-\sqrt{21})$, and $S_3 = \frac{1}{2}(1+\sqrt{21})$.