Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

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

share|improve this question
    
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

1 Answer 1

up vote 5 down vote accepted

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.

share|improve this answer
    
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
    
@user1301428: In that case I guess that it’s a good thing that I mentioned the real mistake. :-) –  Brian M. Scott May 18 '12 at 11:14
    
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
    
@user1301428: Yes, that looks fine. –  Brian M. Scott May 18 '12 at 11:33
    
Thanks for your help again :) –  user1301428 May 18 '12 at 11:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.