2道分解因式&1道XYZ题..高手快来啊

1、a³+b³+c³-3abc
=(a³+b³)+(c³-3abc)
=(a+b)(a²-ab+b²)+c(c²-3ab)
=(a+b)(a²-ab+b²)+c(c²-3ab-a²-b²+a²+b²)
=(a+b)(a²-ab+b²)+c[(c²-a²-2ab-b²)+(a²-ab+b²)]
=(a+b)(a²-ab+b²)+c[c²-(a+b)²]+c(a²-ab+b²)
=(a+b)(a²-ab+b²)+c(a²-ab+b²)+c[c²-(a+b)²]
=(a+b+c)(a²-ab+b²)+c(a+b+c)(c-a-b)
=(a+b+c)[(a²-ab+b²)+c(c-a-b)]
=(a+b+c)(a²+b²+c²-ab-bc-ac)

2、
x³＋3x²＋3x＋2
=(x³＋2x²＋x)＋(x²+2x+1)+1
=x(x²+2x+1)＋(x²+2x+1)+1
=(x+1)(x²+2x+1)+1
=(x+1)(x+1)²+1
=(x+1)³+1
=[(x+1)+1]*[(x+1)²-(x+1)+1]
=(x+2)(x²+x+1)

3、已知x、y、z是实数，且x/(y+z)＋y/(z+x)＋z/(x+y)＝0，
求x²/(y+z)＋y²/(z+x)＋z²/(x+y)的值。

解：可知x+y+z≠0，由已知，得：
1+x/(y+z)