<s>[INST] How to solve system equation $x^3+y^3+z^3=x+y+z, x^2+y^2+z^2=xyz$ for $x,y,z\in\mathbb{R}$? How to solve system equation $\left\{\begin{matrix}x^3+y^3+z^3=x+y+z&\\x^2+y^2+z^2=xyz&\end{matrix}\right.$ , $x,y,z\in\mathbb{R}$ ?
 [/INST] 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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