## Intermediate Algebra (6th Edition)

$\sqrt[6] (y^{5})$
$\sqrt[3] (y^{2})\times\sqrt[6] y=y^{\frac{2}{3}}\times y^{\frac{1}{6}}$ We can further simplify by using the product rule, which holds that $a^{m}\times a^{n}=a^{m+n}$. $y^{\frac{2}{3}}\times y^{\frac{1}{6}}=y^{(\frac{2}{3}+\frac{1}{6})}=y^{(\frac{4}{6}+\frac{1}{6})}=y^{\frac{5}{6}}=\sqrt[6] (y^{5})$