How do you simplify ${(\frac{- x^{4} y^{4} \cdot - x^{- 2} y^{3}}{- x y^{- 1}})}^{4}$ $(\frac { - x ^ { 4} y ^ { 4} \cdot - x ^ { - 2} y ^ { 3} } { - x y ^ { - 1} } ) ^ { 4}$ ?

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How do you simplify ${(\frac{- x^{4} y^{4} \cdot - x^{- 2} y^{3}}{- x y^{- 1}})}^{4}$ $(\frac { - x ^ { 4} y ^ { 4} \cdot - x ^ { - 2} y ^ { 3} } { - x y ^ { - 1} } ) ^ { 4}$ ?

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Answer 1 · 2018-07-24T02:01:58

$x^{4} y^{32}$ $x^4y^32$

Explanation:

${(\frac{- x^{4} y^{4} \cdot - x^{- 2} y^{3}}{- x y^{- 1}})}^{4}$ $((-x^4y^4*-x^(-2)y^3)/(-xy^(-1)))^4$

Numerator:
$- x^{4} y^{4} \cdot - x^{- 2} y^{3}$ $-x^4y^4*-x^(-2)y^3$
= $- x^{4} y^{4} \times - \frac{y^{3}}{x^{2}}$ $-x^4y^4times-y^3/x^2$
= $x^{2} y^{7}$ $x^2y^7$

Denominator:
$- x y^{- 1}$ $-xy^(-1)$
= $- \frac{x}{y}$ $-x/y$

Therefore,
${(\frac{- x^{4} y^{4} \cdot - x^{- 2} y^{3}}{- x y^{- 1}})}^{4}$ $((-x^4y^4*-x^(-2)y^3)/(-xy^(-1)))^4$ becomes ${(\frac{x^{2} y^{7}}{- \frac{x}{y}})}^{4}$ $((x^2y^7)/(-x/y))^4$
= ${((x^{2} y^{7}) \times - \frac{y}{x})}^{4}$ $((x^2y^7)times-y/x)^4$
= ${(- x y^{8})}^{4}$ $(-xy^8)^4$
= $x^{4} y^{32}$ $x^4y^32$

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How do you simplify ${(\frac{- x^{4} y^{4} \cdot - x^{- 2} y^{3}}{- x y^{- 1}})}^{4}$ $(\frac { - x ^ { 4} y ^ { 4} \cdot - x ^ { - 2} y ^ { 3} } { - x y ^ { - 1} } ) ^ { 4}$ ?

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Explanation:

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How do you simplify ${(\frac{- x^{4} y^{4} \cdot - x^{- 2} y^{3}}{- x y^{- 1}})}^{4}$ $(\frac { - x ^ { 4} y ^ { 4} \cdot - x ^ { - 2} y ^ { 3} } { - x y ^ { - 1} } ) ^ { 4}$ ?