Simplify (12x^3-16x^2y+3xy^2+9y^2)(2x^-3y)^-1

$(12 x^{3} - 16 x^{2} y + 3 x y^{2} + 9 y^{2}) {(2 x^{- 3} y)}^{- 1}$

Step 1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$(12 x^{3} - 16 x^{2} y + 3 x y^{2} + 9 y^{2}) {(2 \frac{1}{x^{3}} y)}^{- 1}$

Step 2

Combine $2$ and $\frac{1}{x^{3}}$ .

$(12 x^{3} - 16 x^{2} y + 3 x y^{2} + 9 y^{2}) {(\frac{2}{x^{3}} y)}^{- 1}$

Step 3

Combine $\frac{2}{x^{3}}$ and $y$ .

$(12 x^{3} - 16 x^{2} y + 3 x y^{2} + 9 y^{2}) {(\frac{2 y}{x^{3}})}^{- 1}$

Step 4

Change the sign of the exponent by rewriting the base as its reciprocal.

$(12 x^{3} - 16 x^{2} y + 3 x y^{2} + 9 y^{2}) \frac{x^{3}}{2 y}$

Step 5

Multiply $12 x^{3} - 16 x^{2} y + 3 x y^{2} + 9 y^{2}$ by $\frac{x^{3}}{2 y}$ .

$\frac{(12 x^{3} - 16 x^{2} y + 3 x y^{2} + 9 y^{2}) x^{3}}{2 y}$

Step 6

Reorder factors in $\frac{(12 x^{3} - 16 x^{2} y + 3 x y^{2} + 9 y^{2}) x^{3}}{2 y}$ .

$\frac{x^{3} (12 x^{3} - 16 x^{2} y + 3 x y^{2} + 9 y^{2})}{2 y}$