SOLUTION: Simplify: (x^3-8)/(x^2-3x+9)×(x^3+27)/(x^2+2x+4)×1/(x^2+x-6) Hope this makes sense and that it's the right category to submit it.?

Algebra ->  Numeric Fractions Calculators, Lesson and Practice -> SOLUTION: Simplify: (x^3-8)/(x^2-3x+9)×(x^3+27)/(x^2+2x+4)×1/(x^2+x-6) Hope this makes sense and that it's the right category to submit it.?      Log On


   

Question 888264: Simplify:
(x^3-8)/(x^2-3x+9)×(x^3+27)/(x^2+2x+4)×1/(x^2+x-6)
Hope this makes sense and that it's the right category to submit it.?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
%28x%5E3-8%29%2F%28x%5E2-3x%2B9%29×%28x%5E3%2B27%29%2F%28x%5E2%2B2x%2B4%29×1%2F%28x%5E2%2Bx-6%29
Factor as much as possible
the 1st numerator can be factored as the "difference of cubes"
the 2nd numerator can be factored as the "sum of cubes"
%28%28x-2%29%28x%5E2%2B2x%2B4%29%29%2F%28%28x%5E2-3x%2B9%29%29×%28%28x%2B3%29%28x-3x%2B9%29%29%2F%28x%5E2%2B2x%2B4%29×1%2F%28%28x%2B3%29%28x-2%29%29
Note that we can cancel (x^2+2x+4) and (x^2-3x+9), then we have
(x-2)*(x+3) ×1%2F%28%28x%2B3%29%28x-2%29%29
cancel the last denominator, we are left with just 1
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I do not know what happened yesterday, but I lost it. We can usually correct this using the edit feature, but for some reason the edit feature was not working. Saw that it was working tonight. Hope this helps you, CK