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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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I see some trivial solutions : $(a,b,c) = (1,1,2)$. And note that your system is symmetric under permutations. –  Patrick Da Silva Sep 9 '11 at 3:57
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What have you tried? –  Jack Sep 9 '11 at 3:57
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a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca) is one that you will be requiring from where u will get abc.then substitute one int erms of other 2 to get 2 variable equation which will be easy to solve –  Bhargav Sep 9 '11 at 3:59
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Another maybe interesting remark : your system of equations can be rewritten as $$ \begin{pmatrix} a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \\ \end{pmatrix} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 6 \\ 10 \end{pmatrix}. $$ Note that the 3x3 is a Vandermonde Matrix with determinant $\pm(a-b)(a-c)(b-c)$ (I wrote $\pm$ because I was not sure of the sign). Maybe something nice happens when you invert it or something like that. The solutions found above thus lie in the null space... bleh. Just saying. –  Patrick Da Silva Sep 9 '11 at 4:05
    
I have wondered how you could figure out if a given nonlinear system is actually really simple. For example just by using the newtons identities you can transform a univariate polynomial equation $(x-r_1)(x-r_2)(x-r_3)...=0$ where ($r_1,... $are all known) into a system of $n$ polynomial equations with $n$ variables. How can you see that this is just looking for the roots of a univariate polynomial (and maybe similar problems)? Anyone got any pointers? –  Peter Sheldrick Sep 9 '11 at 5:22

2 Answers 2

up vote 9 down vote accepted

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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Aryabhata solved a very similar problem this way. –  J. M. Sep 9 '11 at 11:20
    
Awesome answer. It shows unicity of the answer found by inspection... and it is highly non-trivial that there was only one solution. +1 –  Patrick Da Silva Sep 11 '11 at 20:35

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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Lol, look at the comments bro. –  Patrick Da Silva Sep 9 '11 at 4:03
    
@Patrick Da Silva: they crossed in etherspace –  Ross Millikan Sep 9 '11 at 4:24
    
From the looks of this, the three solutions already found occur with multiplicity two. Edit: The link doesn't work for some reason; the URL is http://www.wolframalpha.com/input/?i=plot+d^2+%2B+e^2+%2B+%281-d-e%29^2+%3D+1%2‌​C+d^3+%2B+e^3+%2B+%281-d-e%29^3+%3D+1) –  Rahul Sep 9 '11 at 4:31
    
@Rahul Narain: you need to escape the ^ signs to make the link work. Alpha only finds these three solutions –  Ross Millikan Sep 9 '11 at 12:54

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