How do you simplify [(2x^-3 * y^-1) / (4x^-5 * y^-4)]^2 * [(8x^5 * y^6) / (4x^7 * y^4)]^3?

Oct 11, 2017

$2 {x}^{-} 2 {y}^{12} \mathmr{and} \frac{2 {y}^{12}}{x} ^ 2$

Explanation:

${\left[\frac{2 {x}^{-} 3 \cdot {y}^{-} 1}{4 {x}^{-} 5 \cdot {y}^{-} 4}\right]}^{2} \cdot {\left[\frac{8 {x}^{5} \cdot {y}^{6}}{4 {x}^{7} \cdot {y}^{4}}\right]}^{3}$

Simplifying the coefficients first..

${\left[\frac{{\cancel{2}}^{1} {x}^{-} 3 \cdot {y}^{-} 1}{{\cancel{4}}^{2} {x}^{-} 5 \cdot {y}^{-} 4}\right]}^{2} \cdot {\left[\frac{{\cancel{8}}^{2} {x}^{5} \cdot {y}^{6}}{{\cancel{4}}^{1} {x}^{7} \cdot {y}^{4}}\right]}^{3}$

${\left[\frac{{x}^{-} 3 \cdot {y}^{-} 1}{2 {x}^{-} 5 \cdot {y}^{-} 4}\right]}^{2} \cdot {\left[\frac{2 {x}^{5} \cdot {y}^{6}}{{x}^{7} \cdot {y}^{4}}\right]}^{3}$

Using Indices..

Recall; $\Rightarrow {x}^{a} / {x}^{b} = {x}^{a - b}$

${\left[\left(\frac{1}{2} {x}^{- 3 - \left(- 5\right)} \cdot {y}^{- 1 - \left(- 4\right)}\right)\right]}^{2} \cdot {\left[\left(2 {x}^{5 - 7} \cdot {y}^{6 - 4}\right)\right]}^{3}$

Simplifying the indexes..

${\left[\left(\frac{1}{2} {x}^{- 3 + 5} \cdot {y}^{- 1 + 4}\right)\right]}^{2} \cdot {\left[\left(2 {x}^{-} 2 \cdot {y}^{2}\right)\right]}^{3}$

${\left[\left(\frac{1}{2} {x}^{2} \cdot {y}^{3}\right)\right]}^{2} \cdot {\left[\left(2 {x}^{-} 2 \cdot {y}^{2}\right)\right]}^{3}$

Multiplying the indexes to their respective brackets..

$\left({\left(\frac{1}{2}\right)}^{2} {x}^{2 \times 2} \cdot {y}^{3 \times 2}\right) \cdot {2}^{3} {x}^{- 2 \times 3} \cdot {y}^{2 \times 3}$

$\left(\frac{1}{4} {x}^{4} \cdot {y}^{6}\right) \cdot 8 {x}^{-} 6 \cdot {y}^{6}$

$\left(\frac{1}{\cancel{4}} _ 1 {x}^{4} \cdot {y}^{6}\right) \cdot {\cancel{8}}^{2} {x}^{-} 6 \cdot {y}^{6}$

${x}^{4} \cdot {y}^{6} \cdot 2 {x}^{-} 6 \cdot {y}^{6}$

Using Indices..

Recall; $\Rightarrow {x}^{a} \times {x}^{b} = {x}^{a + b}$

$2 {x}^{4 + \left(- 6\right)} \cdot {y}^{6 + 6}$

$2 {x}^{4 - 6} \cdot {y}^{12}$

$2 {x}^{-} 2 {y}^{12} \mathmr{and} \frac{2 {y}^{12}}{x} ^ 2 \to A n s w e r$