The answer is $3x^2 y^2 \sqrt[3]{6y}$
How does $\sqrt[3]{162x^6y^7}$ equal $3x^2 y^2\sqrt[3]{6y}$?
$\endgroup$ 03 Answers
$\begingroup$\begin{align} \sqrt[3]{162x^6y^7}&=\sqrt[3]{27x^6y^6*6y}\\ &=\sqrt[3]{27x^6y^6}\times\sqrt[3]{6y}\\ &=3x^2y^2\sqrt[3]{6y} \end{align}
$\endgroup$ $\begingroup$$162x^6y^7=(3x^2y^2)^3\cdot6y$, such that
$$\sqrt[3]{162x^6y^7}=\sqrt[3]{(3x^2y^2)^3\cdot6y}=\sqrt[3]{(3x^2y^2)^3}\sqrt[3]{6y}=3x^2y^2\sqrt[3]{6y}$$
$\endgroup$ $\begingroup$First note that 162 factors as $2\cdot 3^4$. So collecting our powers of 3 reveals $$\sqrt[3]{162x^6y^7} = \sqrt[3]{3^3 \cdot (x^3)^2\cdot (y^3)^2 \cdot (2\cdot 3 \cdot y)} = 3x^2y^2\sqrt[3]{6y}.$$
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