## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$\sqrt[4]{ 2^{}x^{}y^2}$
Using $a^{-x}=\dfrac{1}{a^x}$ or $\dfrac{1}{a^{-x}}=a^x,$ then \begin{array}{l}\require{cancel} \sqrt[4]{\sqrt[3]{8x^3y^6}} \\\\= \sqrt[4]{ \left( 8x^3y^6 \right)^{1/3}} \\\\= \left( \left( 8x^3y^6 \right)^{1/3} \right)^{1/4} .\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} \left( \left( 8x^3y^6 \right)^{1/3} \right)^{1/4} \\\\= \left( 8x^3y^6 \right)^{\frac{1}{3}\cdot\frac{1}{4}} \\\\= \left( 8x^3y^6 \right)^{\frac{1}{12}} \\\\= \left( 2^3x^3y^6 \right)^{\frac{1}{12}} .\end{array} Using the extended Power Rule of the laws of exponents which is given by $\left( x^my^n \right)^p=x^{mp}y^{np},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \left( 2^3x^3y^6 \right)^{\frac{1}{12}} \\\\= 2^{3\cdot\frac{1}{12}}x^{3\cdot\frac{1}{12}}y^{6\cdot\frac{1}{12}} \\\\= 2^{\frac{1}{4}}x^{\cdot\frac{1}{4}}y^{\frac{2}{4}} .\end{array} Using $x^{m/n}=\sqrt[n]{x^m}=\left(\sqrt[n]{x} \right)^m,$ then \begin{array}{l}\require{cancel} 2^{\frac{1}{4}}x^{\cdot\frac{1}{4}}y^{\frac{2}{4}} \\\\= \sqrt[4]{ 2^{1}x^{1}y^2} \\\\= \sqrt[4]{ 2^{}x^{}y^2} .\end{array}