I'm having trouble factoring this rational expression.

[ECHOSIDE]

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.

 

[ehild]

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

ehild

 

[ECHOSIDE]

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.

 

[Mentallic]

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.

$$\frac{\left(\frac{a^3-b^3}{a^2-2ab+b^2}\right)}{\left(\frac{2a^2+2ab+2b^2}{9a^2-9b^2}\right)}$$

And as you've done, converting it into

$$\frac{a^3-b^3}{a^2-2ab+b^2}\cdot \frac{9a^2-9b^2}{2a^2+2ab+2b^2}$$

Now, your first mistake was an error in factorizing

$$a^3-b^3\neq (a-b)(a^2+2ab+b^2) = (a-b)(a+b)^2$$

It's actually

$$a^3-b^3=(a-b)(a^2+ab+b^2)$$

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.

 

[ECHOSIDE]

This was a huge help. It appears to be accurate. Thank you for taking the time to correct my elementary mistake.