<s>[INST] Solve this system of equations for the value of $a,b$ and $c$ such that $a^3+3ab^2+3ac^2-6abc=1$,$b^3+3ba^2+3bc^2-6abc=1$,$c^3+3cb^2+3ca^2-6abc=1$ Solve this system of equations  for the value of  $a,b$ and $c$.
$$a^3+3ab^2+3ac^2-6abc=1$$
$$b^3+3ba^2+3bc^2-6abc=1$$
$$c^3+3cb^2+3ca^2-6abc=1$$
This is symmetric equation.
My first attempt was,
$a^3+3ab^2+3ac^2=1+6abc$,
$b^3+3ba^2+3bc^2=1+6abc$,
$c^3+3cb^2+3ca^2=1+6abc$,
then, from above three equations,
$a^3+3ab^2+3ac^2=b^3+3ba^2+3bc^2=c^3+3cb^2+3ca^2$
Then, what will I continue?
 [/INST] If we substrac 1. equation -2. equation we get: 
$$(a-b)\Big((a^2+ab+b^2) -3ab+3c^2\Big)=0$$
so if $a\ne b$ we get $$(a-b)^2+3c^2=0 \implies a=b, c=0 $$
a contradiction.
So $a=b$. The same way (2.equation-3.equation) we get $b=c$. S0 $a=b=c =:x$ and now you have to solve: $x^3=1$
</s>