Answer
(a) $-1$
(b) $\frac{1}{729}$
(c) $\frac{1}{x^{5/4}}$
(d) $x^2$
(e) $\frac{b^5}{9}$
(f) $\frac{2x^6}{9y}$
Work Step by Step
(a) $\frac{-2^6}{4^3}=\frac{-2^6}{(2^2)^3}=\frac{-2^6}{2^{3\cdot 2}}=\frac{-2^6}{2^6}=-\frac{2^6}{2^6}=-2^{6-6}=-2^0=-1$
(b) $\frac{(-3)^6}{9^6}=\frac{(-3)^{2\cdot 3}}{9^{6}}=\frac{((-3)^2)^3}{9^6}=\frac{9^3}{9^6}=9^{3-6}=9^{-3}=\frac{1}{9^3}=\frac{1}{729}$
(c) $\frac{1}{\sqrt[4]{x^5}}=\frac{1}{x^{5/4}}$
(d) $\frac{x^3\cdot x^n}{x^{n+1}}=\frac{x^{3+n}}{x^{n+1}}=x^{(3+n)-(n+1)}=x^{2}$
(e) $b^3(3b^{-1})^{-2}=b^3(3^{-2}b^{-1\cdot -2})=b^3(\frac{1}{3^2}\cdot b^2)=\frac{1}{3^2}\cdot b^3\cdot b^2=\frac{1}{9}b^{3+2}=\frac{1}{9}b^5=\frac{b^5}{9}$
(f) $\frac{2x^2y}{(3x^{-2}y)^2}=\frac{2x^2y}{3^2x^{-2\cdot 2}y^2}=\frac{2x^2y}{9x^{-4}y^2}=\frac{2x^{2-(-4)}y^{1-2}}{9}=\frac{2x^6y^{-1}}{9}=\frac{2x^6}{9y}$
