, 2006

Solution:

The answer is (B). By suitable grouping, one writes the cube of $r=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$ 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 $r^{3}=4-3 r$. Among the five proposed numbers, the only one that satisfies the condition is 1.

Second Solution

One easily notices that the product of the two radicals is $-1$, so calling $x$ the first of the two, we ask whether, substituting for $k$ 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 first the answer $k=1$, and it is easy (and a bit astonishing) 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 $y=\sqrt[3]{2-\sqrt{5}}$ we immediately see 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 $s^{3}+3 s-4$ vanishes at 1, is positive for $s>1$, and is negative for $s<1$ (since for $s$ between 0 and 1 also $s^{3}$ is less than 1). Therefore its only real root is 1.