SOLUTION: I have already worked this problem, but think I got confused and mixed up a couple of steps. Could you please provide me with the proper steps so I can evaluate my original work a

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Question 73491This question is from textbook Intermediate Algebra
: I have already worked this problem, but think I got confused and mixed up a couple of steps. Could you please provide me with the proper steps so I can evaluate my original work and see what I did wrong.
Simplify and evaluate:
(2x^-6y^-4z^0/3^-2x^-2y^-8z^5)^3 This question is from textbook Intermediate Algebra

Found 2 solutions by Earlsdon, jim_thompson5910:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Simplify:
First, make all of the negative exponents positive by moving them down or up, as appropriate.
%28%282%2A3%5E2x%5E2y%5E8z%5E0%29%2F%28x%5E6y%5E4z%5E5%29%29%5E3 Replace the z%5E0 with 1 and cancel factors in the numerator and denominator where appropriate.
%28%2818y%5E4%29%2F%28x%5E4z%5E5%29%29%5E3%29 Now we'll cube every thing.
5832%28y%5E12%2F%28x%5E12z%5E15%29%29

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
%282x%5E-6y%5E-4z%5E0%2F3%5E-2x%5E-2y%5E-8z%5E5%29%5E3Start with the original problem
Since %28x%5Ey%29%5Ez=x%5E%28y%2Az%29 the problem can be rewritten as this. Remember each term is raised to the 3rd power, so this rule applies to each term.
Simplify the exponents
Remember to divide variables with exponents, you just subtract the exponents. (eg x%5E2%2Fx%5E1=x%5E%282-1%29=x%5E1 Also x%5E0=1
%288x%5E%28-12%29%29y%5E%2812%29z%5E%28-15%29%2F3%5E%28-6%29%29
Finally anything raised to a negative power is inverted (ie x%5E%28-2%29=1%2Fx%5E2)
%283%5E6%29%288%29y%5E%2812%29%2Fx%5E%2812%29z%5E%2815%29%29
%28729%29%288%29y%5E%2812%29%2Fx%5E%2812%29z%5E%2815%29%29
%285832%29y%5E%2812%29%2Fx%5E%2812%29z%5E%2815%29%29Here it is in simplified form