<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] Let $$a^3+3ab^2+3ac^2-6abc-1=b^3+3a^2b+3bc^2-6abc-1$$ then we get
$$a^3-b^3-3ab(a-b)+3c^2(a-b)=0$$ Now use that $$a^3-b^3=(a-b)(a^2+ab+b^2)$$
Then you will get
$$(a-b)\left((a-b)^2+3c^2)\right)=0$$
Can you finish?
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