1. If $x+y+z=0$, simplify the expression

$$
\frac{x^{7}+y^{7}+z^{7}}{x y z\left(x^{4}+y^{4}+z^{4}\right)}
$$

Hint: first calculate $(x+y)^{4}$ and $(x+y)^{6}$.
1. a) $(x+y)^{4}=x^{4}+4 x^{3} y+6 x^{2} y^{2}+4 x y^{3}+y^{4}$.
b) $(x+y)^{6}=x^{6}+6 x^{5} y+15 x^{4} y^{2}+20 x^{3} y^{3}+15 x^{2} y^{4}+6 x y^{5}+y^{6}$. $\quad 5$ points Now we have

$$
\begin{aligned}
x^{7}+y^{7}+z^{7}= & x^{7}+y^{7}-(x+y)^{7} \\
= & (x+y)\left[x^{6}-x^{5} y+x^{4} y^{2}-x^{3} y^{3}+x^{2} y^{4}-x y^{5}+y^{6}\right. \\
& \left.-\left(x^{6}+6 x^{5} y+15 x^{4} y^{2}+20 x^{3} y^{3}+15 x^{2} y^{4}+6 x y^{5}+y^{6}\right)\right] \\
= & -z\left(-7 x^{5} y-14 x^{4} y^{2}-21 x^{3} y^{3}-14 x^{2} y^{4}-7 x y^{5}\right) \\
= & 7 x y z\left(x^{4}+2 x^{3} y+3 x^{2} y^{2}+2 x y^{3}+y^{4}\right)
\end{aligned}
$$

10 points

$$
x^{4}+y^{4}+z^{4}=x^{4}+y^{4}+(x+y)^{4}=2\left(x^{4}+2 x^{3} y+3 x^{2} y^{2}+2 x y^{3}+y^{4}\right), 5 \text { points }
$$

so by division we get

$$
\frac{x^{7}+y^{7}+z^{7}}{x^{4}+y^{4}+z^{4}}=\frac{7}{2} x y z
$$

and the desired expression is equal to $\frac{7}{2}$.

5 points