Remembering that

$$
\begin{aligned}
(a+b+c)^{3} & =a^{3}+b^{3}+c^{3}+3 a^{2} b+3 a b^{2}+ \\
& +3 a c^{2}+3 a^{2} c+3 b^{2} c+3 b c^{2}+6 a b c
\end{aligned}
$$

Find the solutions to the system of equations

$$
\begin{aligned}
a^{3}+3 a b^{2}+3 a c^{2}-6 a b c & =1 \\
b^{3}+3 b a^{2}+3 b c^{2}-6 a b c & =1 \\
c^{3}+3 c a^{2}+3 c b^{2}-6 a b c & =1
\end{aligned}
$$

#
Solution

Let $A=a^{3}+3 a b^{2}+3 a c^{2}-6 a b c, B=b^{3}+3 b a^{2}+3 b c^{2}-6 a b c$ and $C=c^{3}+3 c a^{2}+3 c b^{2}-6 a b c$. Using the algebraic identity mentioned in the statement, we have

$$
\begin{aligned}
-A+B+C & =(-a+b+c)^{3} \\
A-B+C & =(a-b+c)^{3} \\
A+B-C & =(a+b-c)^{3}
\end{aligned}
$$

Since $A=B=C=1$, it follows that

$$
\begin{aligned}
(-a+b+c)^{3} & =1 \\
(a-b+c)^{3} & =1 \\
(a+b-c)^{3} & =1
\end{aligned}
$$

Therefore

$$
-a+b+c=a-b+c=a+b-c=1
$$

From $2=(-a+b+c)+(a-b+c)=2 c$, it follows that $c=1$. Similarly, we can conclude that $a=b=1$ and that $(a, b, c)=(1,1,1)$ is the only solution to the system.