# Factoring of a fraction, possibly made a mistake

Here is the initial expression and the steps I've made so far, but from the final line I can't go on. Have I made a mistake somewhere?

$$\frac{a+2}{a^3-8}-\frac{1}{a^2-a-2}=\frac{a+2}{(a-2)(a^2+2a+4)}-\frac{1}{(a-2)(a+1)}=\frac{(a+2)(a+1)-(a-2)(a^2+2a+4)}{(a-2)(a^2+2a+4)(a+1)}$$

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How is $a^3+8 = (a-2)(a^2+2a+4)$, $a^3+b^3 = (a+b)(a^2+ab+b^2)$. Further, you might want to check Wolfram Alpha – Inquest May 18 '12 at 9:02
@Nunoxic I've edited the question, I actually meant $a^3-8$ – user1301428 May 18 '12 at 11:13

Corrected to match the corrected question: Your last numerator should be just $(a+2)(a+1)-(a^2+2a+4)$, without the factor of $(a-2)$ in the second term, since you already have a factor of $a-2$ in the second denominator. Then the final fraction simplifies significantly.
I'm sorry, I made a mistake while writing the fraction. I actually really meant $a^-8$ :) I've edited the question. – user1301428 May 18 '12 at 11:13
Thanks for your help. It was a stupid mistake indeed :D As final result I get $1/(a^2+2a+4)(a+1)$, can you confirm it's correct? – user1301428 May 18 '12 at 11:31