Answer
$-2x^4y^3\sqrt[3]{5x^2y}$
Work Step by Step
Factor the radicand so that some factors are perfect cubes:
$\sqrt[3]{-40x^{14}y^{10}}\\
=\sqrt[3]{(-8\cdot 5)(x^{12}\cdot x^2)(y^{9}\cdot y)}$
$=\sqrt[3]{(-8x^{12}y^{9})\cdot 5x^2y}$
With $-8=(-2)^3,(x^4)^3$ and $(y^3)^3$, use the rule $a^nb^nc^n=(abc)^n$ to obtain:
.
$=\sqrt[3]{(-2)^3(x^4)^3(y^3)^3\cdot 5x^2y}\\
=\sqrt[3]{(-2x^4y^3)^3\cdot 5x^2y}$
Separate using the rule $\sqrt[3]{a\cdot b}=\sqrt[3]{a} \cdot \sqrt[3]{b}$ to obtain:.
$=\sqrt[3]{(-2x^4y^3)^3}\sqrt[3]{5x^2y}$
Simplify.
$=-2x^4y^3\sqrt[3]{5x^2y}$
Hence, the correct answer is $-2x^4y^3\sqrt[3]{5x^2y}$.
