(x+y)(y-2z)(2z-x)+2xyz
=(x+y)[(2z-y)x+2z(y-2z)]+2xyz
=(2z-y)\(x^{2}\)+2z(y-2z)x+2xyz+y(2z-y)x+2yz(y-2z)
=(2z-y)\(x^{2}\)-(\(y^{2}\)-6yz+4\(z^{2}\))x+2yz(y-2z)
=(x+y-2z)[x(2z-y)+2yz]
=-(x+y-2z)(xy-2yz-2zx)

2z=pとおくと
(x+y)(y-p)(p-x)
=-(x+y){\(p^{2}\)-(x+y)p+xy}+xyp
=-(x+y)\(p^{2}\)+{(x+y\()^{2}\)+xy}p-xy(x+y) （pの式としてまとめた）
たすきがけで

-(x+y) xy | xy
|
1 -(x+y) | (x+y\()^{2}\)

したがって
与式={-(x+y)p+xy}{p-(x+y)}
=(xy-2zx-2yz)(2z-x-y)・・・（答）