1. ## Adding a Rational Expression with difference of cubes.

I've been stuck on this question for about a half hour. I really don't think the answer I'm getting is correct.

$\frac {(x-y)^2}{x^3-y^3} + \frac {2}{x^2+xy+y^2}$

Is the LCD: (x-y)(x^2+xy+y^2)? I feel like I'm getting thrown off because I'm so use to factoring trinomials when finding the LCD.

The answer I got was and it just doesn't look right:

$\frac {x^2-2xy+y^2+2x-2y}{(x-y)(x^2+xy+y^2)}$

2. The numerator in your answer can be factorised to $2(x-y)+(x-y)^2$

So you can simplify the fraction to $\frac{x-y+2}{x^2+xy+y^2}$

3. Originally Posted by Stroodle
The numerator in your answer can be factorised to $2(x-y)+(x-y)^2$

So you can simplify the fraction to $\frac{x-y+2}{x^2+xy+y^2}$
Thanks, I had forgotten that I could factor out an $x-y$ from $(x-y)^2$ there. My teachers try so much to ingrain in your mind that you can't factor out things when there's a +/- sign separating expressions.

4. Actually, it would have been simpler if you had recognized at the start that $(x- y)(x^2+ xy+ y^2)= x^3- y^3$.

That is, you can immediately reduct the first fraction: $\frac{(x- y)^2}{x^3- y^3}= \frac{(x- y)^2}{(x- y)(x^2+ xy+ y^2}= \frac{x- y}{x^2+ xy+ y^2}$.

Now, $\frac{(x- y)^2}{x^3- y^3}+ \frac{2}{x^2+ xy+ y^2}= \frac{x-y}{x^2+ xy+ y^2}+ \frac{x- y}{x^2+ xy+ y^2}+ \frac{2}{x^2+ xy+ y^2}$ $= \frac{x- y+ 2}{x^2+ xy+ y^2}$.