(a) Verify the identity

$$
(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+3(a+b)(b+c)(c+a)
$$

(b) Solve the system

$$
\left\{\begin{array}{l}
x+y+z=1 \\
x^{2}+y^{2}+z^{2}=1 \\
x^{3}+y^{3}+z^{3}=1
\end{array}\right.
$$
(a) Let's expand $(a+b+c)^{3}$ as $[(a+b)+c]^{3}$.

$$
\begin{aligned}
{[(a+b)+c]^{3} } & =(a+b)^{3}+3(a+b) c(a+b+c)+c^{3} \\
& =a^{3}+b^{3}+3 a b(a+b)+3(a+b) c(a+b+c)+c^{3} \\
& =a^{3}+b^{3}+c^{3}+3(a+b)[a b+c(a+b+c)] \\
& =a^{3}+b^{3}+c^{3}+3(a+b)\left(c^{2}+c(a+b)+a b\right] \\
& =a^{3}+b^{3}+c^{3}+3(a+b)(b+c)(c+a)
\end{aligned}
$$

(b) Using the identity verified in part (a), we get

$$
(x+y+z)^{3}=x^{3}+y^{3}+z^{3}+3(x+y)(y+z)(z+x)
$$

Substituting the values of $x+y+z$ and $x^{3}+y^{3}+z^{3}$, we arrive at

$$
1^{3}=1+3(x+y)(y+z)(z+x)
$$

where $(x+y)(y+z)(z+x)=0$. Thus, $x=-y$ or $y=-z$ or $z=-x$. Since the solutions are symmetric, let's assume $x=-y$. Therefore, from $x+y+z=1$, we get $z=1$ and from $x^{2}+y^{2}+z^{2}=1$ we get $2 x^{2}=0$, or $x=0$. We conclude that the possible solutions are $(0,0,1),(0,1,0)$ and $(1,0,0)$.