SOLUTION: Hi. Could you tell me how I do this problem? (27a^3-8b^3)/(b^2-b-6)*(bc-b-3c+3)/(3ac-2bc-3a+2b)

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Hi. Could you tell me how I do this problem? (27a^3-8b^3)/(b^2-b-6)*(bc-b-3c+3)/(3ac-2bc-3a+2b)      Log On


   

Question 1041309: Hi. Could you tell me how I do this problem?
(27a^3-8b^3)/(b^2-b-6)*(bc-b-3c+3)/(3ac-2bc-3a+2b)
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
%2827a%5E3-8b%5E3%29%2F%28b%5E2-b-6%29%22%D7%22%28bc-b-3c%5E%22%22%2B3%29%2F%283ac%5E%22%22-2bc-3a%2B2b%29
Factor all the numerators and denominators:
27a%5E3-8b%5E3%22%22=%22%22%283a%29%5E3-%282b%29%5E3%22%22=%22%22%283a-2b%29%28%283a%29%5E2%2B%283a%29%282b%29%2B%282b%29%5E2%29%22%22=%22%22%283a-2b%29%289a%5E2%2B6ab%2B4b%5E2%29
b%5E2-b-6%22%22=%22%22%28b-3%29%28b%2B2%29
bc-b-3c%2B3%22%22=%22%22b%28c-1%29-3%28c-1%29%22%22=%22%22%28b-3%29%28c-1%29
3ac-2bc-3a%2B2b%22%22=%22%22c%283a-2b%29-%283a-2b%29%22%22=%22%22%28c-1%29%283a-2b%29

%22%D7%22%28%28b%5E%22%22-3%29%28c%5E%22%22-1%29%29%2F%28%28c%5E%22%22-1%29%283a%5E%22%22-2b%29%29
Cancel the (3a-2b)'s
%22%D7%22
Cancel the (b-3)'s
%22%D7%22
Cancel the (c-1)'s
%22%D7%22
All that left is:
%289a%5E2%2B6ab%2B4b%5E2%29%2F%28b%5E%22%22%2B2%29