I'm given the problem
$$250x^{3}-128y^{3}$$
I know the difference of cubes follows the factored form $$(A-B)(A^{2}+AB+B^{2})$$
Taking the cube root of $250x^{3}$, say A, and the cube root of $128y^{3}$, say B, $\color{blue}A$ should equal $\color{blue}{5x \sqrt[3]2}$ and $\color{red}B$ should equal $\color{red}{4y\sqrt[3]{2}}$, therefore
$$[\color{blue}{(5x\sqrt[3]2)}-\color{red}{(4y\sqrt[3]2)}]\ [\color{blue}{(5x\sqrt[3]2)}^2+\color{blue}{(5x\sqrt[3]2)}\color{red}{(4y\sqrt[3]2)}+\color{red}{(4y\sqrt[3]2)}^2]$$
The answer I keep getting is $$2(5x-4y)(25x^2+20xy+16y^2+\sqrt[3]4)$$ When the answer is supposed to be $$2(5x-4y)(25x^2+20xy+16y^2)$$
How do I get rid of $\sqrt[3]4$ from the solution? The only method I thought of was letting $u=\sqrt[3]2$ and substituting u, then dividing the expression by $u$, giving me the correct answer. Is this sound or did I just get lucky? Here's how it looks:
$$[(5x\sqrt[3]2)-(4y\sqrt[3]2)][(5x\sqrt[3]2)^2+(5x\sqrt[3]2)(4y\sqrt[3]2)+(4y\sqrt[3]2)^2]$$ $$u=\sqrt[3]2 \implies \frac{[(5xu-(4yu)][(5xu)^2+(5xu)(4yu)+(4yu)^2]}{u}$$ $$(5x-4y)((25x^2$+20xy+16y^2)u+u+u)$$ $$u+u+u=\sqrt[3]8=2$$ $$2(5x-4y)(25x^2+20xy+16y^2)$$
HINT
\begin{align*} 250x^{3} - 128y^{3} = 2(125x^{3} - 64y^{3}) = 2[(5x)^{3} - (4y)^{3}] = \ldots \end{align*}