已知X＞Y，求证X^3－Y^3＞X^2Y－XY^2

X^3－Y^3-(X^2Y－XY^2)
=X^2(X-Y)+Y^2(X-Y)
=(X^2+Y^2)(X-Y)
X^2+Y^2>0 X-Y>0

(x-y)(x^2+xy+y^2)>xy(x-y)
因为x>y即x-y>0
所以前式消去(x-y)并移项可得：
x^2+y^2>0
由于x与y不可能同时=0
所以x^2+y^2>0成立

x>y => x-y>0

X^3－Y^3＞X^2Y－XY^2
(x-y)(x^2+xy+y^2)>xy(x-y)
x^2+xy+y^2>xy
所以成立