<s>[INST] 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.
 [/INST] You're given
$$x^3+y^3+z^3=x+y+z \tag{1}\label{eq1A}$$
$$x^2+y^2+z^2=xyz \tag{2}\label{eq2A}$$
As you've shown, you have
$$x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx) \tag{3}\label{eq3A}$$
Next, note
$$(x-y)^2 + (y-z)^2 + (x-z)^2 = 2(x^2 + y^2 + z^2 - xy - yz - zx) \tag{4}\label{eq4A}$$
Since $x,y,z$ are positive values, this means the RHS of \eqref{eq3A} is non-negative (note you can also determine the LHS is non-negative by the Inequality of arithmetic and geometric means which gives $\frac{x^3 + y^3 + z^3}{3} \ge \sqrt[3]{x^3 y^3 z^3} = xyz$), so using this and \eqref{eq2A} gives
$$\begin{equation}\begin{aligned}
x^3 + y^3 + z^3 - 3xyz & \ge 0 \\
x^3 + y^3 + z^3 - 3(x^2 + y^2 + z^2) & \ge 0 \\
x^2(x - 3) + y^2(y - 3) + z^2(z - 3) & \ge 0
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
This shows at least one of $x,y,z$ is $\ge 3$. Due to the symmetry of \eqref{eq1A} and \eqref{eq2A}, WLOG assume $x \le y \le z$, so $z \ge 3$. From \eqref{eq1A}, moving the terms on the right to the left & factoring gives
$$x(x^2 - 1) + y(y^2 - 1) + z(z^2 - 1) = 0 \tag{6}\label{eq6A}$$
With $z \ge 3$, then $z(z^2 - 1) \ge 3(8) = 24$. However, the minimum values of $x(x^2 - 1)$ and $y(y^2 - 1)$ are each $-\frac{2}{3\sqrt{3}} \approx -0.385$ (this comes from $f(a) = a^3 - a$, with $f'(a) = 3a^2 - 1 = 0 \implies a = \frac{1}{\sqrt{3}}$ gives the minimum value of $f(a)$ for $a \gt 0$), so the sum of those $2$ terms is much greater than $-24$. This means the LHS of \eqref{eq6A} can never be $0$ and, thus, the equation never holds. As such, there are no positive, real solutions to \eqref{eq1A} and \eqref{eq2A}.
</s>