## Calculus: Early Transcendentals 8th Edition

(a) $$\frac{x^{2n}\cdot x^{3n-1}}{x^{n+2}}=x^{4n-3}.$$ (b) $$\frac{\sqrt{a\sqrt{b}}}{\sqrt[3]{ab}}=a^\frac{1}{6}b^{-\frac{1}{12}}.$$
(a) To simplify this follow the steps below: $$\frac{x^{2n}\cdot x^{3n-1}}{x^{n+2}}=\frac{x^{2n+(3n-1)}}{x^{n+2}}=\frac{x^{5n-1}}{x^{n+2}}=x^{5n-1-(n+2)}=x^{4n-3}.$$ (b) To simplify the expression, follow the steps below: $$\frac{\sqrt{a\sqrt{b}}}{\sqrt[3]{ab}}=\frac{(a(b)^{\frac{1}{2}})^{\frac{1}{2}}}{(ab)^{\frac{1}{3}}}=\frac{a^{\frac{1}{2}}b^{\frac{1}{2}\cdot\frac{1}{2}}}{a^{\frac{1}{3}}b^{\frac{1}{3}}} = a^{\frac{1}{2}-\frac{1}{3}}b^{\frac{1}{4}-\frac{1}{3}}= a^\frac{1}{6}b^{-\frac{1}{12}}$$