10. What is the value of $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$?
(A) $\sqrt[3]{9-4 \sqrt{5}}$
(B) 1
(C) $\frac{3}{2}$
(D) $\sqrt[3]{4}$
(E) $2 \sqrt[3]{2}$.

## Numerical Answer Problems -8 points
10. The answer is (B). With appropriate factorizations, the cube of $r=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$ can be written as $r^{3}=4+3 \sqrt[3]{2^{2}-5}(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}})=4-3(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}})=4-3 r$. Therefore, $r$ satisfies the condition that $r^{3}=4-3 r$. The only one among the five proposed numbers that satisfies this condition is 1.

Second Solution

It is easily noticed that the product of the two radicals is -1, so if we let $x$ be the first of the two, we ask ourselves if substituting $k$ with one of the five answers, the equation $x-1 / x=k$ has among its solutions precisely $\sqrt[3]{2+\sqrt{5}}$. The equation can be rewritten as $x^{2}-k x-1=0$ and the solutions are $\frac{k \pm \sqrt{k^{2}+4}}{2}$. This expression immediately suggests trying the answer $k=1$ first, and it is easy (and somewhat surprising) to discover that indeed $\left(\frac{1 \pm \sqrt{5}}{2}\right)^{3}=2 \pm \sqrt{5}$.

Third Solution

We have $(x+y)^{3}=x^{3}+y^{3}+3 x y(x+y)$. Taking $x=\sqrt[3]{2+\sqrt{5}}$ and $y=\sqrt[3]{2-\sqrt{5}}$, it is immediately clear that $x^{3}+y^{3}=4$ and $x y=-1$, so $x+y=s$ satisfies the equation $s^{3}=4-3 s$. Now the polynomial $s^{3}+3 s-4$ is zero at 1, positive for $s>1$, and negative for $s<1$ (since for $s$ between 0 and 1, $s^{3}$ is also less than 1). Therefore, its only real root is 1.