## Intermediate Algebra (12th Edition)

$\sqrt[6]{x^{5}}$
$\bf{\text{Solution Outline:}}$ To simplify the given expression, $\sqrt[]{x}\cdot\sqrt[3]{x} ,$ express the radicals as radicals with same indices by finding the $LCD$ of the indices. Once the indices are the same, use the laws of radicals to simplify the expression. $\bf{\text{Solution Details:}}$ The $LCD$ of the indices, $2$ and $3 ,$ is $6$ since it is the lowest number that can be divided exactly by both indices. Multiplying the index by a number to make it equal to the $LCD$ and raising the radicand by the same multiplier results to \begin{array}{l}\require{cancel} \sqrt[2(3)]{x^3}\cdot\sqrt[3(2)]{x^2} \\\\= \sqrt[6]{x^3}\cdot\sqrt[6]{x^2} .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the expression above is equivalent to\begin{array}{l}\require{cancel} \sqrt[6]{x^3(x^2)} .\end{array} Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \sqrt[6]{x^{3+2}} \\\\= \sqrt[6]{x^{5}} .\end{array}