Để đơn giản, Anh viết $x\sim x^2;y\sim y^2;z\sim z^2;a\sim a^2;b\sim b^2.$Ta có:
$\frac{xyz}{ab}+\frac{(x-a)(y-a)(z-a)}{a(a-b)}+\frac{(x-b)(y-b)(z-b)}{b(b-a)}$
$=\frac{xyz}{ab}+\frac{xyz-a(xy+yz+zx)+a^2(x+y+z)-a^3}{a(a-b)}-\frac{xyz-b(xy+yz+zx)+b^2(x+y+z)-b^3}{b(a-b)}$
$=xyz(\frac{1}{ab}+\frac{1}{a(a-b)}-\frac{1}{b(a-b)})-\frac{xy+yz+zx }{a-b}+\frac{xy+yz+zx }{a-b} +\frac{x+y+z}{a-b}(a-b)-\frac{a^2}{a-b}+\frac{b^2}{a-b}$
$=xyz(\frac{a-b+b-a}{ab(a-b)})+0+x+y+z-\frac{a^2-b^2}{a-b}$
$=0+0+x+y+z-\frac{(a+b)(a-b)}{a-b}=x+y+z-a-b.$