Solve this system of equations for the value of $a,b$ and $c$ such that $a^3+3ab^2+3ac^2-6abc=1$,$b^3+3ba^2+3bc^2-6abc=1$,$c^3+3cb^2+3ca^2-6abc=1$

Question

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?

Answer 1

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: $x^3=1$

Answer 2

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?