<s>[INST] find all triplets $(a,b,c)$ such that $ab+bc+ca=1$ and $a^2b+c=b^2c+a=c^2a+b$. Given that $a,b,c \in R$ find all triplets $(a,b,c)$ such that $ab+bc+ca=1$ and $a^2b+c=b^2c+a=c^2a+b$.
Attempt:
Case $I:$ Exactly one of $a,b,c$ is zero. WLOG, let $a=0$. Then we have
$$bc=1$$ and
$$c=b^2c=b$$
Solving we get
$$b=c=\pm 1$$
By the same symmetry we get $6$ triplets viz. $(1,1,0),(-1,-1,0),(1,0,1),(-1,0,-1),(0,1,1),(0,-1,-1)$
Case $II:$ None of $a,b,c $ is Zero.
Let us assume:
$$a^2b+c=b^2c+a=c^2a+b=p$$
Then we have:
$$a^2b^2+bc=bp$$
$$b^2c^2+ac=cp$$
$$c^2a^2+ab=ap$$
By adding all the above equations and
Using the fact that:
$$a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-2abc(a+b+c)=1-2bac(a+b+c)$$
We get
$$2-2abc(a+b+c)=p(a+b+c)$$
$\textbf{ANY HINT PLZ}$
EDIT:
Case $III:$
If $a=b=c \ne 0$ Then we get two more triplets.
$$\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right),\left(\frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\right)$$
 [/INST] The case $a,b,c \neq 0 \implies  c = \dfrac{a(1-ab)}{1-b^2}= \dfrac{ac(a+b)}{1-b^2}\implies a^2+ab=1-b^2\implies a^2+ab+b^2 = 1$. Thus it follows that: $b^2+bc+c^2 = 1 = c^2+ac+a^2\implies a^2+b^2+c^2 = 1$. But Cauchy-Schwarz inequality says: $1 = a^2+b^2+c^2 \ge ab + bc + ac = 1$. So $a = b = c = \pm \dfrac{1}{\sqrt{3}}$, and this is the only solutions for this case. The other cases are treated already.
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