1. Let real numbers $a, b, c$ satisfy $a+b+c=0, a^{3}+b^{3}+c^{3}=0$, find the value of $a^{19}+b^{19}+c^{19}$.
1. Let $a, b, c$ be the three roots of the cubic polynomial $f(x)=t^{3}+a_{1} t^{2}+a_{2} t+a_{3}$. By Newton's formulas, we have $T_{1}=-a_{1}=0$, i.e., $a_{1}=0$. $T_{3}=-a_{1} T_{2}-a_{2} T_{1}-3 a_{3}=-3 a_{3}=0$, i.e., $a_{3}=0$. Therefore, when $k \geqslant 3$, $T_{k}=-a_{2} T_{k-2}$, thus $T_{19}=$ $-a_{2} T_{17}=a^{2} T_{15}=\cdots=-a_{2}^{9} T_{1}=0$. Hence, $a^{19}+b^{19}+c^{19}=0$.