<s>[INST] Find the positive integers $a,b,$ and $c$ s.t. $\sqrt{\sqrt[3]{5} - \sqrt[3]{4}}\times 3 = \sqrt[3]{a}+\sqrt[3]{b}-\sqrt[3]{c}.$ 
Let $a,b,c$ be positive integers such that $\sqrt{\sqrt[3]{5} - \sqrt[3]{4}}\times 3 = \sqrt[3]{a}+\sqrt[3]{b}-\sqrt[3]{c}$. Determine the values of $a, b,$ and $c$.

To solve this simplification problem, it seems that it will be useful to use the difference of cubes formula and/or difference of squares formula as well as some substitutions. I wasn't able to determine a useful method of solving this, however. Using a computer program, I was able to deduce that the desired values $(a,b,c)$ are $(2,20,25)$.
 [/INST] Let $x,y,z\in\mathbb{R}$ such that $x^3 = 5, y^3 = 4$, and $z^3 = 2$. Then $y=z^2, y^2 = 2z$, and $yz = 2.$ We want to simplify $3\sqrt{x-y}$. Note that 
$$\begin{align} 
(x-y)(x+y)^2&=(x-y)(x^2+2xy+y^2)\\
&=x^3-y^3+x^2y-xy^2\\
&=1+(xz)^2-2xz\\
&=(xz-1)^2.
\end{align}$$
Hence 
$$\begin{align}
3\sqrt{x-y} &= 3\cdot\dfrac{(xz-1)(x^2-xy+y^2)}{(x+y)(x^2-xy+y^2)}\\
&=3\cdot\dfrac{x^3z-x^2-x^2(yz)+xy+x(y^2)z-y^2}{x^3+y^3}\\
&=\dfrac{1}{3}(3z-3x^2+3xz^2)=z+xz^2-x^2\\
&=\sqrt[3]{2}+\sqrt[3]{20}-\sqrt[3]{25}\\
&\Rightarrow (a,b,c) = (2,20,25).
\end{align}$$
</s>