Solve the following system of equations

$$
\begin{gathered}
x+y=a \ldots \\
z+t=b \ldots \\
x y=z t \ldots \\
x^{3}+y^{3}+z^{3}+t^{3}=c \ldots
\end{gathered}
$$

Application: $a=10, b=5, c=1080$.
Our equations are symmetric with respect to $x$ and $y$, and also with respect to $z$ and $t$. Since $x+y$ and $z+t$ are known, we can aim to calculate the values of the products $xy$ and $zt$.

It is known that

$$
\begin{aligned}
x^{3}+y^{3} & =(x+y)^{3}-3 x y(x+y) \\
z^{3}+t^{3} & =(z+t)^{3}-3 z t(z+t)
\end{aligned}
$$

Considering this, from (4): $a^{3}+b^{3}-3 a x y-3 b z t=c$.

Since $xy = zt$, it is clear that

$$
xy = zt = \frac{a^{3}+b^{3}-c}{3(a+b)} \cdots
$$

Thus, $x$ and $y$, as well as $z$ and $t$, are the roots of the equations

$$
u^{2}-a u+\frac{a^{3}+b^{3}-c}{3(a+b)}=0 \quad \text { and } \quad v^{2}-b v+\frac{a^{3}+b^{3}-c}{3(a+b)}=0
$$

With the given numerical values,

$$
\begin{aligned}
& \frac{a^{3}+b^{3}-c}{3(a+b)}=\frac{1000+512-1080}{3 \cdot 18}=\frac{432}{54}=8 \\
& x \text { and } y \text { are the roots of } u^{2}-10 u+8=0 \text {; these are } 5 \pm \sqrt{17} \text {, } \\
& z \text { and } t \text { are the roots of } v^{2}-8 v+8=0 \text {; these are } 4 \pm 2 \sqrt{2} \text {. }
\end{aligned}
$$