Solve the following system of equations:

$$
\begin{aligned}
a^{3} x+a^{2} y+a z+u & =0 \\
b^{3} x+b^{2} y+b z+u & =0 \\
c^{3} x+c^{2} y+c z+u & =0 \\
d^{3} x+d^{2} y+d z+u & =1
\end{aligned}
$$
The easiest way to eliminate $u$ is, for example, by subtracting the first equation from the others:

\[
\begin{aligned}
& \left(b^{3}-a^{3}\right) x+\left(b^{2}-a^{2}\right) y+(b-a) z=0 \\
& \left(c^{3}-a^{3}\right) x+\left(c^{2}-a^{2}\right) y+(c-a) z=0 \\
& \left(d^{3}-a^{3}\right) x+\left(d^{2}-a^{2}\right) y+(d-a) z=1
\end{aligned}
\]

If $a$ is different from $b$, $c$, and $d$, then the equations can be simplified by dividing by $(b-a)$, $(c-a)$, and $(d-a)$, respectively, and after simplification, $z$ can be easily eliminated, and then $y$ after further simplifications. The following solution is obtained if $a, b, c$, and $d$ are pairwise distinct:

\[
\begin{aligned}
x & =\frac{1}{(d-a)(d-b)(d-c)}, & y & =-\frac{a+b+c}{(d-a)(d-b)(d-c)} \\
z & =\frac{a b+b c+c a}{(d-a)(d-b)(d-c)}, & u & =-\frac{a b c}{(d-a)(d-b)(d-c)}
\end{aligned}
\]