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How to solve system equation $\left\{\begin{matrix}x^3+y^3+z^3=x+y+z&\\x^2+y^2+z^2=xyz&\end{matrix}\right.$ , $x,y,z\in\mathbb{R}$ ?

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I have tried so many times; but i couldn't find the solution –  Lê Tấn Khang Oct 24 '13 at 15:06

3 Answers 3

We'll using the following notation:

$$s_1 = x + y + z$$ $$s_2 = x^2 + y^2 + z^2$$ $$s_3 = x^3 + y^3 + z^3$$

And we'll express $x,y,z$ as root of the cubic equation $f(x) = ax^3 + bx^2 + cx + d$

From the Vieta's formulas we know that $d = -xyz$

From the condition we have $d = - s_2$ and $s_1 = s_3$

Now use Newton's identitities and we have:

$$as_1 + b = 0$$

$$b = -as_1$$

Now repeat the process:

$$as_2 + bs_1 + 2c = 0$$ $$as_2 + -as_1^2 + 2c = 0$$ $$a(s_2 - s_1^2) + 2c = 0$$ $$2c = -a(s_2 - s_1^2)$$ $$c = \frac{-a(s_2 - s_1^2)}{2}$$

And one last time:

$$as_3 + bs_2 + cs_1 + 3d = 0$$ $$as_1 - as_1s_2 + \frac{-as_1(s_2 - s_1^2)}{2} - 3s_2 = 0$$ $$as_1\left( 1 - s_2 - \frac{s_2 - s_1^2}{2}\right) - 3s_2 = 0$$ $$as_1\left(\frac{-s_2 + s_1^2 + 2 - 2s_2}{2}\right) = 3s_2$$ $$a = \frac{6s_2}{s_1(-s_2 + s_1^2 + 2 - 2s_2)}$$

Now for arbitrary $s_1$ and $s_2$ you should be able to obtain a cubic equation, which roots will satisfy the condition.

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Express all of the terms in terms of the symmetric polynomials $s_1=x+y+z, s_2=xy+yz+zx$ and $s_3=xyz$.

You get $s_1^3-3s_1s_2+3s_3=x^3+y^3+z^3=x+y+z=s_1$ and $s_1^2-2s_2=x^2+y^2+z^2=xyz=s_3$.

Putting the second equation into the first gives $s_1^3-3s_1s_2+3s_1^2-6s_2=s_1$, which simplifies to $s_2=(s_1^3+3s_1^2-s1)/(3s_1+6)$ and $s_3=s_1^2-2s_2$.

Then the solutions to the original equations are the roots of the polynomial $x^3-s_1x^2+s_2x-s_3$ where $s_2$ and $s_3$ are given according to the previous paragraph.

However, for almost all values of $s_1$, 2 of these roots are complex, so these are not real solutions.

In order to get real solutions, you need the discriminant of $x^3-s_1x^2+s_2x-s_3$ to be non-negative. This simplifies to $s_1^2(s_1+2)(648+536s_1+426s_1^2+117s_1^3+12s_1^4-6s1^5-6s1^6+s1^7)$ being non-positive, and the only $s_1$'s for which this is satisfied are $s_1=0$ ($x,y,z=0$) and $-2.16821<=s_1<-2$.

Example non-0 solution are the roots to $100x^3+210x^2-2023x-4487$

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You might use a different unknown for the final equation, say $t$ instead of $x$. We know that the name of the unknown is mostly irrelevant, but since this can confuse readers, it's better sticking to distinct names. –  egreg Oct 25 '13 at 10:04

We can show that there is no solution for $x,y,z>0$. Firstly we have: $$ x^{2}+y^{2}+z^{2}=xyz \implies xyz\geq 3(xyz)^{\frac 23}\\ \implies xyz\geq 27. $$ On the other hand we have: $$ x^{2}+y^{2}+z^{2}=xyz\geq \frac 13(x+y+z)^{2}=\frac 13(x^3+y^3+z^3)^{2}\geq \frac 13(3xyz)^{2}=3(xyz)^2\\ xyz\leq \frac 13 $$ So we have $xyz\leq \frac 13$ and $xyz\geq 27$ which is a contradiction and hence there is no solution for positive $x,y,z$.

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