find a,b,c where $\sqrt{5^{1/3}-4^{1/3}}=\frac{1}{3}(a^{1/3}+b^{1/3}-c^{1/3})$ Is it ramanujan problems? $$\sqrt{5^{1/3}-4^{1/3}}=\frac{1}{3}(a^{1/3}+b^{1/3}-c^{1/3})$$
find $a,b,c$ 
Any helps would be appreciated.
 A: By  squaring of the both sides easy to see that
$$\sqrt{\sqrt[3]5-\sqrt[3]4}=\frac{1}{3}\left(\sqrt[3]2+\sqrt[3]{20}-\sqrt[3]{25}\right)$$
Since $\sqrt[3]2+\sqrt[3]{20}-\sqrt[3]{25}>0$, we need to prove that:
$$9(\sqrt[3]5-\sqrt[3]4)=\sqrt[3]4+\sqrt[3]{400}+\sqrt[3]{625}+2\sqrt[3]{40}-2\sqrt[3]{50}-2\sqrt[3]{500}$$ or
$$9(\sqrt[3]5-\sqrt[3]4)=\sqrt[3]4+2\sqrt[3]{50}+5\sqrt[3]{5}+4\sqrt[3]{5}-2\sqrt[3]{50}-10\sqrt[3]{4},$$
which is obvious.
Yes, it's the Ramanujan's problem.
For example, see here: Denesting radicals like $\sqrt[3]{\sqrt[3]{2} - 1}$
A: Note that 
\begin{eqnarray*}
\left( \sqrt[3]{\alpha}+\sqrt[3]{\alpha^2 \beta }-\sqrt[3]{\beta^2}\right) ^2 &=& \color{blue}{\sqrt[3]{\alpha^2}}+\color{red}{\alpha\sqrt[3]{\alpha \beta^2}} +\beta\sqrt[3]{\beta}+2 \alpha\sqrt[3]{\beta} \color{red}{-2 \sqrt[3]{\alpha\beta^2}} \color{blue}{-2 \beta\sqrt[3]{\alpha^2}}  \\
&=& \color{blue}{(1-2\beta)\sqrt[3]{\alpha^2}}+\color{red}{(\alpha-2)\sqrt[3]{\alpha \beta^2}} +(2 \alpha+\beta)\sqrt[3]{\beta}
\end{eqnarray*}
Now by inspection a solution can be obtained by choosing $\alpha=2$ and $\beta=5$, giving $a=2,b=20,c=25$.
