Solve the equations and find $x,y,z$ Given:
$$x^3+y^3+z^3=x+y+z$$
And:
$$x^2+y^2+z^2=xyz$$ 
Find all real and positive solutions to these equations, if any.
So most probably, we'll factorise in this way:
$$x^3+y^3+z^3-3xyz=x+y+z-3xyz
\rightarrow (x+y+z)(x^2+y^2+z^2-xy-yz-zx)=(x+y+z)-3(x^2+y^2+z^2)$$
But I'm having trouble in proceeding further.
Kindly help.

Denote by $\sigma_i$ the elementary symmetric functions and by $p_r$ the power sums of the nonnegative variables $x$, $y$, $z$. We are told that
$$p_3=\sigma_1,\qquad p_2=\sigma_3\ .\tag{1}$$
Of course $x=y=z=0$ is a solution of $(1)$. We therefore may assume $\sigma_1>0$ in the sequel. Using the inequalities $$\left({x+y+z\over3}\right)^2\leq{x^2+y^2+z^2\over3},\qquad \bigl(\sigma_3\bigr)^{1/3}\leq{\sigma_1\over3}$$
we find 
$${1\over3}\sigma_1^2\leq p_2=\sigma_3\leq{1\over27}\sigma_1^3\ ,$$
so that  $\sigma_1\geq9$.  On the other hand
$${x^3+y^3+z^3\over3}\geq\left({x+y+z\over3}\right)^3$$
implies
$$\sigma_1=p_3\geq{1\over9}\sigma_1^3\ ,$$
hence $\sigma_1\leq3$. It follows that the given equations $(1)$ cannot be solved with nonnegative $x$, $y$, $z$ other than $(0,0,0)$.
