Solve the system of equations

$$
\left\{\begin{array}{l}
a^{3}+3 a b^{2}+3 a c^{2}-6 a b c=1 \\
b^{3}+3 b a^{2}+3 b c^{2}-6 a b c=1 \\
c^{3}+3 c a^{2}+3 c b^{2}-6 a b c=1
\end{array}\right.
$$

in real numbers.

Answer: $a=1, b=1, c=1$.
Denoting the left hand sides of the given equations as $A, B$ and $C$, the following equalities can easily be seen to hold:

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

Hence, the system of equations given in the problem is equivalent to the following one:

$$
\left\{\begin{array}{c}
(-a+b+c)^{3}=1 \\
(a-b+c)^{3}=1 \\
(a+b-c)^{3}=1
\end{array}\right.
$$

which gives

$$
\left\{\begin{array}{c}
-a+b+c=1 \\
a-b+c=1 \\
a+b-c=1
\end{array} .\right.
$$

The unique solution of this system is $(a, b, c)=(1,1,1)$.