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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asked Sep 9, 2011 at 3:42
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1$\begingroup$ I see some trivial solutions : $(a,b,c) = (1,1,2)$. And note that your system is symmetric under permutations. $\endgroup$– Patrick Da SilvaCommented Sep 9, 2011 at 3:57
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1$\begingroup$ 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 $\endgroup$– BhargavCommented Sep 9, 2011 at 3:59
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2$\begingroup$ 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. $\endgroup$– Patrick Da SilvaCommented Sep 9, 2011 at 4:05
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$\begingroup$ The question Three-variable system of simultaneous equations deals with a very similar system of equations. Only the numbers on the RHS are different. $\endgroup$– Martin SleziakCommented Jun 21, 2013 at 7:12
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2 Answers
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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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$\begingroup$ Aryabhata solved a very similar problem this way. $\endgroup$ Commented Sep 9, 2011 at 11:20
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$\begingroup$ Awesome answer. It shows unicity of the answer found by inspection... and it is highly non-trivial that there was only one solution. +1 $\endgroup$ Commented Sep 11, 2011 at 20:35
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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.
answered Sep 9, 2011 at 4:01
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$\begingroup$ Lol, look at the comments bro. $\endgroup$ Commented Sep 9, 2011 at 4:03
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$\begingroup$ @Patrick Da Silva: they crossed in etherspace $\endgroup$ Commented Sep 9, 2011 at 4:24
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$\begingroup$ 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%2C+d^3+%2B+e^3+%2B+%281-d-e%29^3+%3D+1)$\endgroup$– user856Commented Sep 9, 2011 at 4:31 -
$\begingroup$ @Rahul Narain: you need to escape the ^ signs to make the link work. Alpha only finds these three solutions $\endgroup$ Commented Sep 9, 2011 at 12:54