10. Solve the system of equations: $\left\{\begin{array}{l}x+a y+a^{2} z+a^{3}=0 \\ x+b y+b^{2} z+b^{3}=0 \\ x+c y+c^{2} z+c^{3}=0\end{array}\right.$
10. Solution Observe the shape characteristics of the three equations in the system, which inspires the construction of a cubic equation $t^{3}+z t^{2}+y t+x=0$, where $a, b, c$ are its three roots, and $a, b, c$ are also the roots of the equation $(t-a)(t-b)(t-c)=0$.
Expanding the latter equation yields $t^{3}-(a+b+c) t^{2}+(a b+b c+c a) t-a b c=0$.
By comparing the coefficients of these two cubic equations, we can obtain the solution to the original system of equations as $x=-a b c, y=a b+b c+c a, z=-(a+b+c)$.