algebra problem?

Question

(x^-3)(9xy^2)^3

Answer 1

do those pointy arrows mean "to the power of"? if so then....

i got something like 729x^-9y^6 ....... i could be totally off though.... sounds like you distribute the power of 3 first , then multiply by x to the negative third power .... hope that helps-maybe more answers will help choose what is right, im not sure how to distribute the x^-3 throughout at this point (x^-3)729x^3y^6


EDIT: there we go- at the end you the x to the engative third cancels with the x to the third leaving you with 729y^6 :) wow that was a workout

Answer 2

oaky.. whenever you have a negative exponent you always put the whole term under 1. so for the first term you would get 1/ (x^3) right? okay for the next term.. you will put all the terms in the parentheses to the power of 3. So.. 9^3, x^3, (y^2)^3. Whenever you have the power of a term then power again.. on the outside.. you will want to multiply the exponents... but if you are multiplying two similar terms together.. like x^3 times X^5, then you would want to add the exponents. so back to the problem.. you will then get 729 (x^3) (y^6). but you notice the 1/ (x^3) and (x^3) right? so you can cross both of them off.. you will be left with 729 (y^6) right? if you have any quesitons, feel free to email me at [e-mail]

Answer 3

Formula:

x ^ -y = 1/x^y

In your formula:
(x^-3) = 1/x^3
(1/x^3) * (9xy^2)^3 = (1/x^3) * (9xy^6)
(1/x^3) * (9xy^6) = (9xy^6 / x^3)

Good luck.

Answer 4

The effect of multiplying (1/x^3) times x^3 is to remove x from the problem. So your answer will be 729 y^6

Answer 5

729y^6

(x^-3)=1/x3
9^3=729
(729x^3y^6)/x^3=729y^6
the x^3 on top cancels out the one on the bottom leaving you with one on the bottom which leaves you with
729y^6

Answer 6

(x^-3)(9xy^2)^3
= (x^-3)(9x^3y^6)
=9x^-2(3y^6)
=3y^6/9x^2
=y^6 / 3x^2