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

$3x^3y^4 \sqrt[3] {2x^2y}$
Apply product rule: $\sqrt[n] p \sqrt[n] q=\sqrt[n]{pq}$ Here, $n$ refers as index. $\sqrt[3] {6x^7y} \sqrt[3] {9x^4y^{12}}=\sqrt[3] {(6x^7y) (9x^4y^{12})}= \sqrt[3]{54x^{11}y^{13}}$ The radical $\sqrt[3]{54x^{11}y^{13}}$ can be further simplified by using product rule again. Such as: $\sqrt[3]{54x^{11}y^{13}}=\sqrt[3] {(3x^3y^4)^3 (2x^2y)}=\sqrt[3] {(3x^3y^4)^3} \sqrt[3]{ (2x^2y)}=3x^3y^4 \sqrt[3] {2x^2y}$