# How do you simplify (2x^3y^2)^2(2x^-1y^4)^3 / (2^1x^2y^-3)^-1?

Jun 25, 2015

Expand the expression to get rid of the brackets, and then remove any common factors.

#### Explanation:

${\left(2 {x}^{3} {y}^{2}\right)}^{2}$ = $4 {x}^{6} {y}^{4}$

${\left(2 {x}^{-} 1 {y}^{4}\right)}^{3} = 8 {x}^{-} 3 {y}^{12}$
${\left({2}^{1} {x}^{2} {y}^{-} 3\right)}^{-} 1$ = ${2}^{-} 1 {x}^{-} 2 {y}^{3}$

$4 {x}^{6} {y}^{4} \cdot 8 {x}^{-} 3 {y}^{12} / \left({2}^{-} 1 {x}^{-} 2 {y}^{3}\right)$ = $32 {x}^{3} {y}^{16} \cdot 2 {x}^{2} {y}^{-} 3$

= $64 {x}^{5} {y}^{13}$