I'm having trouble factoring this rational expression.

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  • #1
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Homework Statement



((a^3-b^3)/(a^2-2ab+b^2))/((2a^2+2ab+2b^2)/(9a^2-9b^2))

Not using complex number system.
Not concerned with domain.
Find the quotient and put it in simplest terms.

Homework Equations





The Attempt at a Solution



Too many to transcribe. Apparently I'm missing a legal operation somewhere. Wolfram|Alpha gives ((9)(a+b))/2, and I am able to reach equivalent answers, but not that simplest one.

Thank you for any help you can provide.
 
Last edited:

Answers and Replies

  • #2
ehild
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Show some of your work please. The solution is very simple, just factorize everything.

ehild
 
  • #3
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my work so far

Thank you for your response. Here is my work so far.

((a^3-b^3)/(a^2-2ab+b^2))/((2a^2+2ab+2b^2)/(9a^2-9b^2))

((a^3-b^3)/(a^2-2ab+b^2))*((9a^2-9b^2)/(2a^2+2ab+2b^2))

((a-b)(a^2+2ab+b^2)/(a^2-2ab+b^2))*((9a^2-9b^2)/(2a^2+2ab+2b^2))

((a-b)(a^2+2ab+b^2)/(a^2-2ab+b^2))*((9)(a^2-b^2)/(2a^2+2ab+2b^2))

((a-b)(a^2+2ab+b^2)/(a^2-2ab+b^2))*((9)(a+b)(a-b)/(2a^2+2ab+b^2))

((a-b)(a^2+2ab+b^2)/(a^2-2ab+b^2))*((9)(a+b)(a-b)/2(a^2+ab+b^2))

((a-b)(a+b)(a+b)/(a-b)(a-b))*((9)(a+b)(a-b)/2(a^2+ab+b^2)

Edit: The last step should be (((a-b)(a+b)(a+b))/((a-b)(a-b)))*(((9)(a+b)(a-b))/(2(a^2+ab+b^2))). I missed a few critical parentheses in there. This should translate properly.

The trouble here is with the quantity (a^2+ab+b^2). Does that factor? Did I arrive there in error?

Thank you for any help you can provide.
 
Last edited:
  • #4
Mentallic
Homework Helper
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Homework Statement



((a^3-b^3)/(a^2-2ab+b^2))/((2a^2+2ab+2b^2)/(9a^2-9b^2))
I'm going to convert this to latex because it's just a pain to read in text format.

[tex]\frac{\left(\frac{a^3-b^3}{a^2-2ab+b^2}\right)}{\left(\frac{2a^2+2ab+2b^2}{9a^2-9b^2}\right)}[/tex]

And as you've done, converting it into

[tex]\frac{a^3-b^3}{a^2-2ab+b^2}\cdot \frac{9a^2-9b^2}{2a^2+2ab+2b^2}[/tex]

Now, your first mistake was an error in factorizing

[tex]a^3-b^3\neq (a-b)(a^2+2ab+b^2) = (a-b)(a+b)^2[/tex]

It's actually

[tex]a^3-b^3=(a-b)(a^2+ab+b^2)[/tex]

And so the second factor can't be factorized further. This should help you along because the same factor can be found in the denominator of the second fraction.
 
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  • #5
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This was a huge help. It appears to be accurate. Thank you for taking the time to correct my elementary mistake.
 

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