令x+y=z，则f(x+y,x-y)=f(z,z-2y)=yz(z-y)，再令z-2y=t,则f(z,z-2y)=f(z,t)=1/2(z-t)z[z-1/2(z-t)]=1/4(z²-t²)z 故f(x,y)=1/4(x²-y²)x
换元法，f(x,y)=x*(x²-y²)/4
f(x+y,x-y)=x²y+xy²=xy(x+y)=1/4*[(x+y)²-(x-y)²](x+y)
则f(x,y)=1/4*(x²-y²)x
设u=x+y，v=x-y
则x=(u+v)/2，y=(u-v)/2
∴f(u，v)=xy(x+y)
=(u²-v²)/4·u
=(u³-uv²)/4
∴f(x，y)=(x³-xy²)/4
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