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How would one go about analytically solving a system of non-linear equations of the form:

$a + b + c = 4$

$a^2 + b^2 + c^2 = 6$

$a^3 + b^3 + c^3 = 10$

Thanks!

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    The question [Three-variable system of simultaneous equations](http://math.stackexchange.com/questions/27394/three-variable-system-of-simultaneous-equations/) deals with a very similar system of equations. Only the numbers on the RHS are different.2013-06-21

2 Answers 2

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Hint: Newton's identities.

$\begin{align*}a+b+c&=4\\ ab+bc+ca=\frac12((a+b+c)^2-(a^2+b^2+c^2))&=\frac12(4^2-6)=5\\abc=\frac13((ab+bc+ca-a^2-b^2-c^2)(a+b+c)+(a^3+b^3+c^3))&=\frac13((5-6)\cdot 4+10)=2\end{align*}$

Remember that $a,\ b,\ c$ are the three roots of a polynomial $P(t)=(t-a)(t-b)(t-c)=t^3-(a+b+c)t^2+(ab+bc+ca)t-abc$.

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    Awesome answer. It shows unicity of the answer found by inspection... and it is highly non-trivial that there was only one solution. +12011-09-11
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One solution is $1,1,2$ by inspection.

If we rewrite it in $d=a-1, e=b-1, f=c-1$, the equations become

$d+e+f=1$

$d^2+e^2+f^2=1$

$d^3+e^3+f^3=1$

and all the variables, if real, must be in $[-1,1]$. I would expect six solutions from the product of the degrees and have found three. So it is natural to assume two variables are equal to make three more, but the first two equations then yield $\frac{2}{3},\frac{2}{3},\frac{-1}{3}$, which does not satisfy the last.

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    @Rahul Narain: you need to escape the ^ signs to make the link work. Alpha only finds these three solutions2011-09-09