2. Let real numbers $x, y, z, w$ satisfy $x+y+z+w=x^{7}+y^{7}+z^{7}+w^{7}=0$, find the value of $w(w+x)(w+y)(w+z)$.
(IMO - 26 Shortlist)
2. Let $x, y, z, w$ be the four roots of the quartic polynomial $f(x)=t^{4}+a_{1} t^{3}+a_{2} t^{2}+a_{3} t+a_{4}$, and let $T_{n}=x^{n}+y^{n}+z^{n}+w^{n}$. Then $T_{1}=-a_{1}=0$, i.e., $a_{1}=0$. $T_{2}=-2 a_{2}$, $T_{3}=-3 a_{3}$, $T_{4}=2 a_{2}^{2}-4 a_{4}$, $T_{5}=5 a_{2} a_{3}$, $T_{7}=7 a_{3}(a_{4}-a_{2}^{2})=0$, so $a_{4}-a_{2}^{2}=0$ or $a_{3}=0$. (i) If $a_{4}-a_{2}^{2}=0$, then $T_{4}=-2 a_{2}^{2} \leqslant 0$, but $T_{4}=x^{4}+y^{4}+z^{4}+w^{4} \geqslant 0$, thus $T_{4}=0$, i.e., $x=y=z=w=0$, in this case, $w(w+x)(w+y)(w+z)=0$. (ii) If $a_{3}=0$, then $f(x)=(t-x)(t-y)(t-z)(t-w)=t^{4}+a_{2} t^{2}+a_{4}$. Therefore, the four roots of the polynomial $f(t)$ must consist of two pairs of numbers that are opposites. In this case, $w(w+x)(w+y)(w+z)=0$. Hence, the required value is always 0.