How do we compute $\sqrt[3]{x_1} +\sqrt[3]{x_2} $ using the fact that $x_1 + x_2 = 4 , x_1x_2 = -1$?

Question

Given quadratic equation

$$x^2 -4x-1 = 0$$

How do we compute $\sqrt[3]{x_1} +\sqrt[3]{x_2} $ using the fact that $x_1 + x_2 = 4 , x_1x_2 = -1$?

Regards

Answer 1

Note that: $(\sqrt[3]{x_1}+\sqrt[3]{x_2})^3=(x_1+x_2)+3\sqrt[3]{x_1x_2}(\sqrt[3]{x_1}+\sqrt[3]{x_2})$ Putting $t= \sqrt[3]{x_1}+\sqrt[3]{x_2}$ yields: $t^3+3t-4=0$. The last cubic equation has one real root $t=1$ which gives the needed result.

Answer 2

Using $a^3+b^3= (a+b)(a^2-(ab)+b^2)$:

$(x_1+x_2)=$

$(x_1^{1/3}+x_2^{1/3})(x_1^{2/3} -(x_1x_2)^{1/3}+x_2^{2/3}).$

Set $X = x_1^{1/3}+x_2^{1/3}$

$4=X(x_1^{2/3}+x_2^{2/3}+1).$

Express $x_1^{2/3}+x_2^{2/3}$ in terms of $X$:

$X^2= (x_1^{1/3}+x_2^{1/3})^2=$

$x_1^{2/3}+x_2^{2/3} +2x_1^{1/3}x_2^{1/3}$;

$X^2+2= x_1^{2/3}+x_2^{2/3}.$

Finally:

$4=X(X^2+3)$.

$X^3+3X -4=0.$

$X=1$ is a solution, factoring:

$(X-1)(X^2 +X+4)=0$.

Any other solutions?