Total number of real triplets of $(x,y,z)$ in $x^3+y^3+z^3=x+y+z$ and $x^2+y^2+z^2=xyz$
$\bf{My\; Try::}$ Let $x+y+z=a$ and $xy+yz+zx=b\;,$
Then $(x+y+z)^2-2(xy+yz+zx)=xyz\implies a^2-2b=xyz$
And using $\displaystyle x^3+y^3+z^3-3xyz=(x+y+z)\left[x^2+y^2+z^2-(xy+yz+zx)\right]$
So we get $a-3xyz=a[xyz-b]\implies a+ab=(a+3)xyz$
So we get $\displaystyle \frac{a+ab}{a+3} = a^2-2b$
Now how can I solve it after that, Help required, Thanks.
Assume that $x+y+z=a$ and $xyz=b$. We have: $$ e_2 = xy+xz+yz = \frac{a^2-b}{2} $$ and we may consider that: $$ p(t) = (t-x)(t-y)(t-z) = t^3 - e_1 t^2 + e_2 t^2 - e_3 = t^3 - at^2 + \frac{a^2-b}{2}t - b. $$ On the other hand, for any $w\in\{x,y,z\}$ we have $p(w)=0$ or $$ w^3 = e_1 w^2 - e_2 w + e_3,\tag{1} $$ hence by summing $(1)$ over $w\in\{x,y,z\}$ we get: $$ b=x^3+y^3+z^3 = a b - \frac{a^2-b}{2}a+3b \tag{2}$$ from which $b=\frac{a^3}{4+3a}$. The problem boils down to understanding which values of $a$ grant that $$ p(t) = t^3 - at^2 + \frac{(a+2)a^2}{4+3a}\,t-\frac{a^3}{4+3a}\tag{3} $$ has three real roots, i.e. a non-negative discriminant. That cannot happen if $a>-3$, but happens for sure if $a$ belongs to some sub-interval of $\mathbb{R}^-$, so we have an infinite number of triplets fulfilling the given constraints.