# Algebra for RMO

Problem: Prove that

$\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$ is a rational number.

Proof:

Let $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$. We then have

$x - \sqrt[3]{2+\sqrt{5}} - \sqrt[3]{2-\sqrt{5}}=0$

We know that $a+b+c=0$ implies that $a^{3}+b^{3}+c^{3}=3abc$, so we obtain

$x^{3}-(2+\sqrt{5})-(2-\sqrt{5})=3x\sqrt[3]{(2+\sqrt{5})(2-\sqrt{5})}$

or, $x^{3}+3x-4=0$

Clearly, out of the roots of this equation is $x=1$ and the other two roots satisfy the equation $x^{2}+x+4=0$ which has no real solutions. (This equation can be found by polynomial division). Since $\sqrt[3]{2+\sqrt{5}}+\sqrt{2-\sqrt{5}}$ is a real number, it follows that

$\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1$, which is a rational number.

More later,

Nalin Pithwa

## One thought on “Algebra for RMO”

1. Sachit

We can also cube the number(which is to be proved rational) , to get the same cubic.

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