Algebra 2 Common Core

$\dfrac{\sqrt[3]{150ab^2c}}{5a}$
Multiplying the numerator and the denominator by an expression equal to $1$ which will make the denominator a perfect power of the index, the given expression, $\dfrac{\sqrt[3]{12ab^3c^2}}{\sqrt[3]{10a^3bc}} ,$ is equivalent to \begin{align*} & =\dfrac{\sqrt[3]{12ab^3c^2}}{\sqrt[3]{10a^3bc}}\cdot\dfrac{\sqrt[3]{10^2b^2c^2}}{\sqrt[3]{10^2b^2c^2}} \\\\&= \dfrac{\sqrt[3]{12ab^3c^2(10^2b^2c^2)}}{\sqrt[3]{10^3a^3b^3c^3}} \\\\&= \dfrac{\sqrt[3]{1200ab^5c^4}}{\sqrt[3]{(10abc)^3}} \\\\&= \dfrac{\sqrt[3]{1200ab^5c^4}}{10abc} .\end{align*} Extracting the factors that are perfect powers of the index, the expression above is equivalent to \begin{align*}\require{cancel} & =\dfrac{\sqrt[3]{8b^3c^3\cdot150ab^2c}}{10abc} \\\\&= \dfrac{\sqrt[3]{(2bc)^3\cdot150ab^2c}}{10abc} \\\\&= \dfrac{2bc\sqrt[3]{150ab^2c}}{10abc} \\\\&= \dfrac{\cancel2\cancel{bc}\sqrt[3]{150ab^2c}}{\cancel{10}^5a\cancel{bc}} \\\\&= \dfrac{\sqrt[3]{150ab^2c}}{5a} .\end{align*} Hence, the simplified form of the given expression is $\dfrac{\sqrt[3]{150ab^2c}}{5a}$.