Track / Stage 5 / 189 of 400 #789 of 1964

Problem 789

AIME late

Algebra Difficulty 5.4 Prove it

Prove the equality $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $u=\sqrt[3]{2+\sqrt{5}}, v=\sqrt[3]{2-\sqrt{5}}$ . Then $u^{3}+v^{3}=4, u v=\sqrt[3]{4-5}=-1$ . From the equality $(u+v)^{3}$ $=u^{3}+v^{3}+3 u v(u+v)$ it is clear that $u+v$ is a root of the equation $x^{3}+3 x-4=0$ . This equation has an obvious root $x=1$ , and it has no other roots (see problem $\underline{61252}$ a).

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.