Is this factorization correct?

Question

I have this problem: $$\sqrt[3]{8x^5}-\sqrt[3]{27x^8}$$

The result I get is: $$x\sqrt[3]{x^2}(2-3x)$$ But I don't know if I factored it correctly because the answer is: $$\sqrt[3]{x^2}(2x-3x^2) $$

Answer 1

Your answer is correct: your factorization is in fact more complete, since you factored out $x$ from the sum $\;(2x - 3x^2)$: $$2x - 3x^2 = x(2 - 3x).$$

Indeed: $$x\sqrt[3]{x^2}(2-3x) = \sqrt[3]{x^2}\cdot x(2 - 3x) = \sqrt[3]{x^2}(2x - 3x^2)$$

Answer 2

Your result is equivalent to the one you bring as an answer, just distribute the $x$ in front of the cube root into the parentheses.