(x+y+z)2-x2-y2-z2= 2(xy+yz+zx)
<=>(x+y+z)2-x2-y2-z2-2(xy+yz+zx)=0
<=>(x+y+z)2-x2-y2-z2-2xy-2yz-2zx=0
<=>(x+y+z)2-(x2+y2+z2+2xy+2yz+2zx)=0
<=>(x+y+z)2-[(x2+2xy+y2)+(2yz+2zx)+z2]=0
<=>(x+y+z)2-[(x+y)2+2.(x+y).z+z2]=0
<=>(x+y+z)2-(x+y+z)2=0
<=>0=0 (luôn đúng với mọi x,y,z)
Vậy (x+y+z)2-x2-y2-z2= 2(xy+yz+zx) với mọi x,y,z