How do you simplify $(\frac { 2x y \cdot 2y x ^ { 0} } { ( - 2x y ^ { 4} ) ^ { 2} } ) ^ { - 3}$ ?

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Answer 1 · 2017-09-11T05:38:38

$x^3y^18$

A few things about fractions and exponents we need to keep in mind:

1) $(a/b)^-x=(b/a)^x$

2) $a^x/a^y=a^(x-y)$

3) $a^0=1$

4) $(ab)^x=a^xb^x$

5) $(a^x)^y=a^(x*y)$

So, with that, we have the following:

$((2xy*2yx^0)/(-2xy^4)^2)^-3$

$((-2xy^4)^2/(2xy*2yx^0))^3larr$ using property 1

$((4x^2y^8)/(2xy*2y))^3larr$ property 4 on top and 3 on the bottom

$((4x^2y^8)/(4xy^2))^3larr$ multiplying out the bottom

Now we can divide the top by the bottom using property 2

$((cancel(4)x^cancel(2)y^cancel(8))/(cancel(4)cancel(x)cancel(y^2)))^3$

As you can see, all of the bottom cancels out leaving us with:

$(xy^6)^3$

Now we just use property 5 to get us:

$x^3y^(6*3)$

$x^3y^18$

How do you simplify (\frac { 2x y \cdot 2y x ^ { 0} } { ( - 2x y ^ { 4} ) ^ { 2} } ) ^ { - 3}(\frac { 2x y \cdot 2y x ^ { 0} } { ( - 2x y ^ { 4} ) ^ { 2} } ) ^ { - 3}?