SOLUTION: (x^3-1)/(x^2+1) divided by (9x^2+9x+9)/(x^2-x). THis problem is giving me a real hard time. I cannot seem to factor the 9x^2+9x+9 to anything less that 9(x^2+x+1) which does not ca

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: (x^3-1)/(x^2+1) divided by (9x^2+9x+9)/(x^2-x). THis problem is giving me a real hard time. I cannot seem to factor the 9x^2+9x+9 to anything less that 9(x^2+x+1) which does not ca      Log On


   

Question 53255This question is from textbook Elementary and Intermediate Algebra
: (x^3-1)/(x^2+1) divided by (9x^2+9x+9)/(x^2-x). THis problem is giving me a real hard time. I cannot seem to factor the 9x^2+9x+9 to anything less that 9(x^2+x+1) which does not cancel anything out. Am I doing something wrong? This question is from textbook Elementary and Intermediate Algebra

Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
First flip the second fraction over and multiply:
%28%28x%5E3-1%29%2F%28x%5E2%2B1%29%29%2A%28%28x%5E2-x%29%2F%289x%5E2%2B9x%2B9%29%29
Then factor. The numerator in the first fraction is a difference between two cubes can be factored:
a%5E3-b%5E3=%28a-b%29%28a%5E2%2Bab%2Bb%5E2%29
in your case a=x and b=1.

%28x%28x-1%29%5E2%28x%5E2%2Bx%2B1%29%29%2F%289%28x%5E2%2B1%29%28x%5E2%2Bx%2B1%29%29
x%5E2%2Bx%2B1cancels and we have:
%28x%28x-1%29%5E2%29%2F%289%28x%5E2%2B1%29%29