anonymous
  • anonymous
(-3w^3 x^-6)^3 Write your answer using only positive exponents.
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[x^{-a} = \frac{1}{x^a}\]\[(xy)^m = x^my^m\]
anonymous
  • anonymous
Please help? I still dont understand? ):
jim_thompson5910
  • jim_thompson5910
how far did you get gabriel72alv?

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anonymous
  • anonymous
I couldnt figure out how to put it in the equation I needed to solve it? ):
jim_thompson5910
  • jim_thompson5910
I'll do a similar example (just with different numbers) \[\Large \left(-2w^{5}x^{-8}\right)^{4}\] \[\Large \left(-2\right)^{4}\left(w^{5}\right)^{4}\left(x^{-8}\right)^{4}\] \[\Large \left((-2)^{1}\right)^{4}\left(w^{5}\right)^{4}\left(x^{-8}\right)^{4}\] \[\Large (-2)^{1*4}w^{5*4}x^{-8*4}\] \[\Large (-2)^{4}w^{20}x^{-32}\] \[\Large (-2)^{4}w^{20}\frac{1}{x^{32}}\] \[\Large 16w^{20}\frac{1}{x^{32}}\] \[\Large \frac{16w^{20}}{x^{32}}\]
anonymous
  • anonymous
So would I get (-3w^3/x^6)^3 ?
jim_thompson5910
  • jim_thompson5910
Keep going. So far, so good.
anonymous
  • anonymous
Theres still another step? o:
jim_thompson5910
  • jim_thompson5910
Yes there are more steps.
anonymous
  • anonymous
How would you work it out to get to the next step though? I understand that.
jim_thompson5910
  • jim_thompson5910
You would apply the outermost exponent to each term inside
jim_thompson5910
  • jim_thompson5910
In other words, you would cube each piece.
anonymous
  • anonymous
So basically would it be, (-3w^3/x^6) times (-3w^3/x^6) times (-3w^3/x^6) ?
jim_thompson5910
  • jim_thompson5910
Yes pretty much
jim_thompson5910
  • jim_thompson5910
So that explains why, for instance, in the denominator you will have x^6*x^6*x^6 = x^(6+6+6) = x^(18)
jim_thompson5910
  • jim_thompson5910
A shortcut is to use this rule \[\Large (x^m)^n = x^{m*n}\]
anonymous
  • anonymous
So would I get for my answer to that is -27w^9/x^18 ?
jim_thompson5910
  • jim_thompson5910
You are 100% correct. Nice work.
anonymous
  • anonymous
Nicee! Haa, So thats my answer? xD
jim_thompson5910
  • jim_thompson5910
Yes it is.
anonymous
  • anonymous
Awesome, Thank you so much jim_thompson5910 ! (:
jim_thompson5910
  • jim_thompson5910
you're welcome

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