Evaluating $\sqrt[3]{x_1}+\sqrt[3]{x_2}$

Question

Consider the following quadratic equation:

$$x^2-4x-1 = 0 \tag{1}$$

Whose solutions are denoted by $x_1, x_2\in \mathbb{R}$. Then by using the fact that $x_1+x_2 = 4$ and $x_1x_2 = -1$, how can we evaluate $\sqrt[3]{x_1}+\sqrt[3]{x_2}$?

$$\begin{align}\biggr(\sqrt[3]{x_1}+\sqrt[3]{x_2}\biggr)^3 &= x_1+3(\sqrt[3]{x_1})^2\sqrt[3]{x_2}+3(\sqrt[3]{x_1})(\sqrt[3]{x_2})^2 + x_2\\\ &= (x_1+x_2) + 3(\sqrt[3]{x_1})^2(\sqrt[3]{x_2})+3(\sqrt[3]{x_1})(\sqrt[3]{x_2})^2 \\\ &= 4+3\biggr((\sqrt[3]{x_1})^2(\sqrt[3]{x_2})+(\sqrt[3]{x_1})(\sqrt[3]{x_2})^2\biggr) \\\ &= 4+3\biggr((\sqrt[3]{x_1})\sqrt[3]{x_1x_2} +(\sqrt[3]{x_2})\sqrt[3]{x_1x_2}\biggr) \\\ &= 4-3\biggr(\sqrt[3]{x_1}+\sqrt[3]{x_2}\biggr) \tag{2}\end{align}$$

However, I am not sure where this would lead us. Could I hear your feedback, if possible?

Answer 1

If you mark $t=\sqrt[3]{x_1}+\sqrt[3]{x_2}$ you have to solve $$t^3 = 4-3t$$ which shouldn't be hard (using say rational root theorem).

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Or if $y_i =\sqrt[3]{x_i}$ then you need $y_1+y_2$ knowing $y_1y_1 =-1$ and $y_1^3+y_2^3=4$. From last one you have $$4=(y_1+y_2)(y_1^2-y_1y_2+y_2^2) = t(t^2-3y_1y_2) = t(t^2+3)$$

so you get the same equation.

Answer 2

Alt. hint: $\,$ the roots of the quadratic are $\,x_{1,2} = 2 \pm \sqrt{5}\,$, and $\,\sqrt[3]{x_{1,2}}$ $= \sqrt[3]{2 \pm \sqrt{5}}$ $= \frac{1}{2}\left(1 \pm \sqrt{5}\right)\,$, see Denesting Phi, Denesting Cube Roots, or Generalized denesting formula for $\sqrt[3]{A+B\sqrt{C}}$.