Answer
Practice 5 (Solution)
$(\,\frac{4x^3}{3y^{-1}})\,^3(\,\frac{y^{-2}}{3x^{-1}})\,^{-1}$
=$\frac{(\,4x^3)\,^3}{(\,3y^{-1})\,^3}\cdot\frac{(\,y^{-2})\,^{-1}}{(\,3x^{-1})\,^{-1}}$ (Power of a quotient)
=$\frac{4^3x^9y^2}{3^3y^{-3}\cdot3^{-1}x^1}$ (Power rule)
=$\frac{64x^{9-1}y^{2-(-3)}}{3^{3+(-1)}}$ (Quotient rule, Product rule)
=$\frac{64x^8y^5}{9}$
Work Step by Step
Practice 5 (Solution)
$(\,\frac{4x^3}{3y^{-1}})\,^3(\,\frac{y^{-2}}{3x^{-1}})\,^{-1}$
=$\frac{(\,4x^3)\,^3}{(\,3y^{-1})\,^3}\cdot\frac{(\,y^{-2})\,^{-1}}{(\,3x^{-1})\,^{-1}}$ (Power of a quotient)
=$\frac{4^3x^9y^2}{3^3y^{-3}\cdot3^{-1}x^1}$ (Power rule)
=$\frac{64x^{9-1}y^{2-(-3)}}{3^{3+(-1)}}$ (Quotient rule, Product rule)
=$\frac{64x^8y^5}{9}$
