Evaluation of $\sqrt[3]{40+11\sqrt{13}}+\sqrt[3]{40-11\sqrt{13}}$

Question

Evaluate $$\sqrt[3]{40+11\sqrt{13}}+\sqrt[3]{40-11\sqrt{13}}$$

The solution is $5$.

Suppose $\sqrt[3]{40+11\sqrt{13}}=A, \sqrt[3]{40-11\sqrt{13}}=B$

We have $$A^3+B^3=80, A^3-B^3=22\sqrt{13}$$

Two unknowns, two equations, so we should be able to solve for $A+B$ (what we want).

How can I do that?

Answer 1

Hint 1

$$A^3+B^3=(A+B)((A+B)^2-3AB)$$

Hint 2

$$AB=3$$

Answer 2

Note that $$\sqrt[3]{40+11\sqrt{13}}=\frac{5+\sqrt{13}}{2}$$ $$\sqrt[3]{40-11\sqrt{13}}=\frac{5-\sqrt{13}}{2}$$

Answer 3

Let $x = \sqrt[3]{40 + 11\sqrt{13}} + \sqrt[3]{40 - 11\sqrt{13}}$. Then we have that \begin{align*} x^{3} = 80 + 9x & \Longleftrightarrow x^{3} - 9x - 80 = 0\\\\ & \Longleftrightarrow (x^{3} - 25x) + (16x - 80) = 0\\\\ & \Longleftrightarrow x(x^{2} - 25) + 16(x - 5) = 0\\\\ & \Longleftrightarrow x(x + 5)(x - 5) + 16(x - 5) = 0\\\\ & \Longleftrightarrow (x^{2} + 5x + 16)(x - 5) = 0\\\\ & \Longleftrightarrow x = 5 \end{align*}

and we are done.

Hopefully this helps !