# I'm having trouble factoring this rational expression.

1. May 7, 2013

### ECHOSIDE

1. The problem statement, all variables and given/known data

((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.

2. Relevant equations

3. 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.

Last edited: May 7, 2013
2. May 7, 2013

### ehild

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

ehild

3. May 7, 2013

### 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?

Last edited: May 7, 2013
4. May 7, 2013

### Mentallic

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.

5. May 24, 2013

### ECHOSIDE

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