## gabriel72alv one year ago (-3w^3 x^-6)^3 Write your answer using only positive exponents.

1. RolyPoly

$x^{-a} = \frac{1}{x^a}$$(xy)^m = x^my^m$

2. gabriel72alv

3. jim_thompson5910

how far did you get gabriel72alv?

4. gabriel72alv

I couldnt figure out how to put it in the equation I needed to solve it? ):

5. jim_thompson5910

I'll do a similar example (just with different numbers) $\Large \left(-2w^{5}x^{-8}\right)^{4}$ $\Large \left(-2\right)^{4}\left(w^{5}\right)^{4}\left(x^{-8}\right)^{4}$ $\Large \left((-2)^{1}\right)^{4}\left(w^{5}\right)^{4}\left(x^{-8}\right)^{4}$ $\Large (-2)^{1*4}w^{5*4}x^{-8*4}$ $\Large (-2)^{4}w^{20}x^{-32}$ $\Large (-2)^{4}w^{20}\frac{1}{x^{32}}$ $\Large 16w^{20}\frac{1}{x^{32}}$ $\Large \frac{16w^{20}}{x^{32}}$

6. gabriel72alv

So would I get (-3w^3/x^6)^3 ?

7. jim_thompson5910

Keep going. So far, so good.

8. gabriel72alv

Theres still another step? o:

9. jim_thompson5910

Yes there are more steps.

10. gabriel72alv

How would you work it out to get to the next step though? I understand that.

11. jim_thompson5910

You would apply the outermost exponent to each term inside

12. jim_thompson5910

In other words, you would cube each piece.

13. gabriel72alv

So basically would it be, (-3w^3/x^6) times (-3w^3/x^6) times (-3w^3/x^6) ?

14. jim_thompson5910

Yes pretty much

15. 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)

16. jim_thompson5910

A shortcut is to use this rule $\Large (x^m)^n = x^{m*n}$

17. gabriel72alv

So would I get for my answer to that is -27w^9/x^18 ?

18. jim_thompson5910

You are 100% correct. Nice work.

19. gabriel72alv

Nicee! Haa, So thats my answer? xD

20. jim_thompson5910

Yes it is.

21. gabriel72alv

Awesome, Thank you so much jim_thompson5910 ! (:

22. jim_thompson5910

you're welcome