How do you simplify $\frac { x ^ { - 8} ( x ^ { - 2} y ^ { 2} ) ^ { 2} } { ( 4x ^ { - 4} ) ^ { 2} y ^ { 2} }$ ?

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Answer 1 · 2018-04-23T00:47:17

$(x^-8(x^-2y^2)^2)/((4x^-4)^2y^2)=y^2/(16x^4)$

$(x^-8(x^-2y^2)^2)/((4x^-4)^2y^2)$

Let's loose the parenthesis first. Remember that

$(x^m)^n=x^(mn)$ .

$(x^-8(x^-2y^2)^2)/((4x^-4)^2y^2)= (x^-8x^-4y^4)/(16x^-8y^2)$

Now cancel. Remember that

$(x^m)/(x^n)=x^(m-n)$ .

$(x^-8x^-4y^4)/(16x^-8y^2)=(x^-4y^2)/16$

Now express everything with positive exponents. Remember that

$x^-m=1/x^m$

$(x^-4y^2)/16=y^2/(16x^4)$

How do you simplify \frac { x ^ { - 8} ( x ^ { - 2} y ^ { 2} ) ^ { 2} } { ( 4x ^ { - 4} ) ^ { 2} y ^ { 2} }\frac { x ^ { - 8} ( x ^ { - 2} y ^ { 2} ) ^ { 2} } { ( 4x ^ { - 4} ) ^ { 2} y ^ { 2} }?