A simple math identity about one

Prove that

$$\frac{bc}{(a-b)(a-c)}+ \frac{ca}{(b-c)(b-a)}+\frac{ab}{(c-a)(c-b)}=1$$

given $a \not = b\not =c$ and $a \not= c$.

Proof:

Let $x = a-b$ and $y = b-c$ and $z = a-c$, then we will have the following:

$$\frac{bc}{(a-b) (a-c)} + \frac{ca}{(b-c)(b-a)} + \frac{ab}{(c-a)(c-b)} = \frac{bc}{xz} - \frac{ca}{xy} + \frac{ab}{yz} \\ = \frac{bc y - ca z + ab x}{xyz}$$

Since $x =a -b$ and $y = b-c$ so $x + y =a-c = z$.

and so our original formula will be

$$\frac{bcy -ca(x+y) +abx}{xyz} = \frac{c(b-a)y+a(b-c)x}{xyz} \\= \frac{(a-c)xy}{xyz} = 1$$,

which proves the identity