## Intermediate Algebra for College Students (7th Edition)

$2y^2z \sqrt[4]{2x^3y^3z}$
Simplify . $\sqrt[4]{32x^3y^{11}z^5}$ As per the product rule, we have $\sqrt[n] {pq}=\sqrt[n] {p}\sqrt[n] {q}$ $\sqrt[4]{32x^3y^{11}z^5}=\sqrt[4] {32}\sqrt[4] {x^3}\sqrt[4] {y^{11}} \sqrt[4] {z^5}$ Thus, $=\sqrt[4] {2^4}\sqrt[4] {(y^2)^4}\sqrt[4] {z^4} \sqrt[4]{2x^3y^3z}$ or, $=\sqrt[4]{(2y^2z)^4} \sqrt[4]{2x^3y^3z}$ Hence, the above exponent in radical form can be written as: or, $2y^2z \sqrt[4]{2x^3y^3z}$