How do you simplify $((16x^2y^-1)/(4x^0y^-4z))^-2$ ?

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Answer 1 · 2017-04-14T17:29:44

$(16z^2)/(256x^4y^8)$

Recall: $(a/b)^-2 = (b/a)^2$

(Flip the fraction and the index changes sign.)

$((16x^2y^-1)/(4x^0y^-4z))^-2 = ((4x^0y^-4z)/(16x^2y^-1))^2$

Now simplify the fraction inside the bracket.

$((4x^0y^-4z)/(16x^2y^-1))^2 = ((4(1)yz)/(16x^2y^4))^2$

$((4z)/(16x^2y^3))^2" "larr$ now square all the factors.

$=(16z^2)/(256x^4y^8)$

How do you simplify ((16x^2y^-1)/(4x^0y^-4z))^-2((16x^2y^-1)/(4x^0y^-4z))^-2?