How do you simplify ${[\frac{x y^{6}}{x^{3} y}]}^{- 2}$ $[(xy^6)/(x^3y)]^-2$ ?

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Answer 1 · 2018-03-25T20:56:44

$\Rightarrow \frac{x^{4}}{y^{10}}$ $=> x^4/y^10$

We start with:

$\Rightarrow {[\frac{x y^{6}}{x^{3} y}]}^{- 2}$ $=>[(xy^6)/(x^3y)]^-2$

We multiply all powers by $- 2$ $-2$ :

$\Rightarrow \frac{x^{- 2} y^{6 (- 2)}}{x^{3 (- 2)} y^{- 2}}$ $=>(x^(-2)y^(6(-2)))/(x^(3(-2))y^(-2))$

Simplifying powers:

$\Rightarrow \frac{x^{- 2} y^{- 12}}{x^{- 6} y^{- 2}}$ $=>(x^-2 y^-12)/(x^-6 y ^-2)$

Division of terms with powers requires subtraction of powers:

$\Rightarrow x^{- 2 - (- 6)} y^{- 12 - (- 2)}$ $=> x^(-2 - (-6)) y^(-12 - (-2))$

Simplifying:

$\Rightarrow x^{4} y^{- 10}$ $=>x^4y^-10$

We can get rid of negative powers by moving $y$ $y$ to the denominator:

$\Rightarrow \frac{x^{4}}{y^{10}}$ $=> x^4/y^10$

How do you simplify [xy6x3y]−2[(xy^6)/(x^3y)]^-2?