How do you simplify #(w^2/w^4)^-3#?

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Answer 1 · 2016-09-01T10:48:55

We have that

#(w^2/w^4)^-3=(w^4/w^2)^3=(w^2)^3=w^6#

Answer 2 · 2016-09-01T12:01:54

#w^6#

There are 3 laws of indices which apply here.

#x^m/x^n = x^(m-n)," "x^-m = 1/x^m, " "(x^a y^b)^m = x^(am)y^(bm)#

It does not matter which law you apply first.

#(w^2/w^4)^-3 = (1/w^2)^-3 larr "simplify in the bracket"#

#(1/w^2)^-3 = (w^2)^(+3)larr "invert the fraction"#

#(w^2)^(+3) =w^6 " "larr "power law"#
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OR

#(w^2/w^4)^-3 = (w^-6/w^-12) " "larr "power law"#

#(w^-6/w^-12) = (w^12/w^6) " "larr "invert the fraction"#

#w^12/w^6 = w^6" "larr " simplify"#