1. Let $a, b, c \in \mathbf{R}$. If
$$
a+b+c=0, a^{2}+b^{2}+c^{2}=6 \text {, }
$$

then $a^{4}+b^{4}+c^{4}=$ $\qquad$ .
$-1.18$
$$
\begin{array}{l}
\text { Given } a+b+c=0, a^{2}+b^{2}+c^{2}=6 \\
\Rightarrow b c+c a+a b \\
\quad=\frac{1}{2}\left((a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)\right) \\
\quad=-3 .
\end{array}
$$
$$
\begin{aligned}
\text { Therefore } & a^{4}+b^{4}+c^{4} \\
= & \left(a^{2}+b^{2}+c^{2}\right)^{2}-2(b c+c a+a b)^{2}+ \\
& 4 a b c(a+b+c) \\
= & 18 .
\end{aligned}
$$