[ARML 1997]

Relevant wiki: Rational Numbers - Problem Solving

$\large \sqrt[3]{\sqrt[3]{2}-1} = \sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}.$

Let $x=\sqrt[3]{2}-1$ and $y=\sqrt[3]{2}$ . Then $y^3=2$ , $x=\sqrt[3]{y}-1$ Note that

$1=y^3-1=(y-1)(y^2+y+1)$ and

$y^2+y+1=\frac{3y^2+3y+3}{3}=\frac{y^3+3y^2+3y+1}{3}=\frac{(y+1)^3}{3}$ which implies that

$x^3=y-1=\frac{1}{y^2+y+1}=\frac{3}{(y+1)^3}$

or

$x=\frac{\sqrt[3]{3}}{y+1}.......(1)$

on the other hand

$3=y^3+1=(y+1)(y^2-y+1)$

from which it follows that

$\frac{1}{y+1}=\frac{y^2-y+1}{y^3+1}=\frac{y^2-y+1}{3}........(2)$

Combine $(1)$ and $(2)$ , We obtain

$x=\sqrt[3]{\frac{1}{9}}(\sqrt[3]{4}-\sqrt[3]{2}+1)$

Consequently

$(a,b,c)=(\frac{4}{9},\frac{-2}{9},\frac{1}{9})$ is a desired triple.