Three, to be factored into a product of rational integral expressions in $x$ and $y$

$$
x^{7}+x^{6} y+x^{5} y^{2}+x^{4} y^{3}+x^{3} y^{4}+x^{2} y^{5}+x y^{6}+y^{7}
$$
I. Solution. The given polynomial $\frac{x^{8}-y^{8}}{x-y}$ is equivalent to the following. However,

$$
\begin{gathered}
\frac{x^{8}-y^{8}}{x-y}=\frac{\left(x^{4}-y^{4}\right)\left(x^{4}+y^{4}\right)}{x-y}=\frac{\left(x^{2}-y^{2}\right)\left(x^{2}+y^{2}\right)\left(x^{4}+y^{4}\right)}{x-y}= \\
=(x+y)\left(x^{2}+y^{2}\right)\left(x^{4}+y^{4}\right)
\end{gathered}
$$

Miklós Hajnal (Izr. g. VI. o. Bp.)

II. Solution. If we substitute $x=-y$ into the given expression, the value of the expression becomes zero. This implies that it is divisible by $(x+y)$ and the quotient is

$$
x^{6}+x^{4} y^{2}+x^{2} y^{4}+y^{6}=x^{4}\left(x^{2}+y^{2}\right)+y^{4}\left(x^{2}+y^{2}\right)=\left(x^{4}+y^{4}\right)\left(x^{2}+y^{2}\right) .
$$

György Sommer (Dobó István g. VI. o. Eger.)