Surd Problem 2

Algebra Level 5

$x$ , $y$ and $z$ are rational numbers such that

$\sqrt[3]{\sqrt[3]{2}-1} = \sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}$

If $x+y+z$ can be expressed as $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers,

find $m+n$ .

1 solution

Hugh Sir
Nov 26, 2014

Let $p = \sqrt[3]{\sqrt[3]{2}-1}$ and $q = \sqrt[3]{2}$ .

Then $q^{3} = 2$ and $p = \sqrt[3]{q-1}$ .

Note that $1 = q^{3}-1 = (q-1)(q^{2}+q+1)$ ,

and $q^{2}+q+1 = \frac{3q^{2}+3q+2+1}{3} = \frac{q^{3}+3q^{2}+3q+1}{3} = \frac{(q+1)^{3}}{3}$ .

Then $p^{3} = q-1 = \frac{1}{q^{2}+q+1} = \frac{3}{(q+1)^{3}}$ .

So $p = \frac{\sqrt[3]{3}}{q+1}$ .

On the other hand, $3 = q^{3}+1 = (q+1)(q^{2}-q+1)$ .

Then $\frac{1}{q+1} = \frac{q^{2}-q+1}{3} = \frac{\sqrt[3]{4}-\sqrt[3]{2}+1}{3}$ .

So $p = \frac{\sqrt[3]{3}}{q+1} = \frac{\sqrt[3]{3}}{3}(\sqrt[3]{4}-\sqrt[3]{2}+1) = \sqrt[3]{\frac{1}{9}}(\sqrt[3]{4}+\sqrt[3]{-2}+1)$ .

Hence, $\sqrt[3]{\sqrt[3]{2}-1} = \sqrt[3]{\frac{4}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{1}{9}} = \sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}$ .

$x+y+z = \frac{4}{9}-\frac{2}{9}+\frac{1}{9} = \frac{1}{3} = \frac{m}{n}$ .

Consequently, $m+n = 1+3 = 4$ .