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I've been trying to help my son with what seems an innocent enough equation, but I have to admit, I'm struggling and walking in circles - the question is:

If $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=4$, what is $abc$?

First I multiply both sides with the denominators:

$a(a+c)(a+b)+b(b+c)(a+b)+c(b+c)(a+c)=4(a+b)(a+b)(b+c)$

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$a^3+b^3+c^3=3(abc+a^2b+ab^2+b^2c+a^2c+ac^2+bc^2)$

From here I begin to run in circles - substituting $x=abc$ and expressing $a$, $b$ and $c$ in terms of $x$, for example, ends up after a while back at the beginning. So what is the trick - or is there no simple solution?

(Sorry about the tag - I couldn't find anything more suitable for such an elementary question)

Edit

The suggested duplicate is not a full answer, I think. My question concerns the same equation, but the question is different, and the answers are only partial. I will accept the answer given by Rhys Hughes, since it answers my question and gives a good explanation of how he reached it.

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    $\begingroup$ Related question: math.stackexchange.com/questions/2192461/… $\endgroup$
    – Robert Z
    Feb 6, 2021 at 8:47
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    $\begingroup$ You could choose $a=0, b=1$ and compute $c=2 \pm \sqrt{3}$. $\endgroup$
    – copper.hat
    Feb 6, 2021 at 8:48
  • $\begingroup$ This question is usually for integers. Then $abc$ can only attain very few values, too. $\endgroup$ Feb 6, 2021 at 9:22

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$abc$ is not fixed in general. You can pick $a=b=\alpha$ and arrive at $$\frac{2\alpha}{\alpha+c}+\frac{c}{2\alpha}=4\implies c=\alpha\bigg(\frac{7\pm\sqrt{65}}{2}\bigg)$$Yielding $abc=\alpha^3\bigg(\frac{7\pm\sqrt{65}}{2}\bigg)$.

In this way we may have the value of $abc$ span $\Bbb R$ since for each $y\in \Bbb R$; $\alpha=\bigg(\frac{2y}{7\pm\sqrt {65}}\bigg)^\frac 13\implies abc=y$

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