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I have the following fraction:

$\frac{a^3-8}{a^2+2a+4}$

Because the numerator is the difference of two cubes, I've factored it like this: $(a-2)(a^2+8a+64)$.

The denumerator does not have natural roots, it would be factored in the following way: $((x-(1+(i)\sqrt{3})(x-(1-(i)\sqrt{3}))$.

My question is: can I simplify this fraction in another way without using complex numbers?

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The numerator is $(a-2)(a^2+2a+4)$, from the rule $a\pm b=(a\pm b)(a^2\mp ab+b^2)$. The quadratic term cancels with the denominator, and you get a final result of $a-2$. You can verify this by multiplying $a-2$ by the denominator, which you will see is similar to the telescoping sum of a geometric series. –  bgins May 2 '12 at 12:22

1 Answer 1

up vote 2 down vote accepted

The numerator is $(a-2)(a^2+2a+4)$, from the rule $a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2)$. The quadratic term cancels with the denominator, and you get a final result of $a-2$. You can verify this by multiplying $a-2$ by the denominator, which you will see is similar to the telescoping sum of a geometric series.

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LOL stupid mistake. You are right, thanks :) –  user1301428 May 2 '12 at 12:30
    
You might want to change $a \pm b$ with $a^3 \pm b^3$ ;) I would edit this myself but I can't because the edit is too short. –  user1301428 May 2 '12 at 12:34
    
me too :-) thanks –  bgins May 2 '12 at 15:14

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