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I am here again to ask a question about an exercise I saw around but i'm having a lot of trouble with. I know the answer is 3abc, but as in many of my questions, I am interested in the why and how.

Given $a+b+c=0$, simplify:

$$\frac{(a^3-abc)^3+(c^3-abc)^3+(b^3-abc)^3}{(c^2-ab)(b^2-ac)(a^2-bc)}$$

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What did you try? –  Pedro Tamaroff Mar 24 '13 at 23:31
    
I tried $$a^3+b^3+c^3=3abc$$, but i think that is going to get the equation bigger instead of simplifying it. Also, expanding it, i get nothing. –  chubakueno Mar 24 '13 at 23:35

3 Answers 3

up vote 9 down vote accepted

Your idea that $a^3+b^3+c^3=3abc$ is a good one.

Consider your numerator $x^3+y^3+z^3$, where $x = a^3-abc$ and so forth, try to see what $x+y+z$ is, and what the relationship between $x^3+y^3+z^3$ and $3xyz$ is.

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How beautiful it gets now! Thank you, i think this is the most elegant solution. –  chubakueno Mar 25 '13 at 0:03

If you just sub in $c=-a-b$ then each factor of the denominator reduces to $(a^2+ab+b^2)$.

Each term of the numerator has a factor that is one of these denominator factors as well. For instance the first term from the numerator is $a^3(a^2-bc)^3$, which is $a^3(a^2+ab+b^2)^3$.

So we have $$\begin{align} \frac{a^3(a^2+ab+b^2)^3+c^3(a^2+ab+b^2)^3+b^3(a^2+ab+b^2)^3}{(a^2+ab+b^2)^3} \end{align} $$

Now provided $a^2+ab+b^2\neq0$ (which is guaranteed the case if all our numbers are real), we have $$a^3+b^3+c^3$$

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You can factor e.g. $a^3 - a b c = a(a^2 - b c)$, which matches a factor in the denominator. This suggests a sum of fractions.

You can always write out the numerator and denominator. The similarity of the terms gives hope that there will be lots of cancellation.

Try replacing simple values, like $a = b = c$ and see what happens.

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Thanks for the interest, but I have tried that and failed. According to wolframalpha.com, the numerator is equivalent to $$3abc (a^2+a b+b^2)^3$$ and the denominator is $$(a^2+a b+b^2)^3$$, but I can't understand how did it came to that conclusion. –  chubakueno Mar 24 '13 at 23:44
    
Ok, I replaced a=-b-c and i get the denominator, but the numerator is still a mistery... –  chubakueno Mar 24 '13 at 23:57
    
@chubakueno, neither can I. There will be terms $a^9 + b^9 + c^9$ in the numerator, I don't see how to get rid of them. Unless substituting $a = -b - c$ helps somehow here. But that gives another sort of mess... –  vonbrand Mar 25 '13 at 0:05
    
Thanks to the enlightment @Sanchez gave me, I got the solution. Review his answer if you have any doubts vonbrand.Thanks for your help, happy mathing, and goodbye!. –  chubakueno Mar 25 '13 at 0:10
    
@chubakueno, looks fine. –  vonbrand Mar 25 '13 at 0:15

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