Prove the following expression Prove that if $a,b,c$ are not equal:
$$\dfrac {(x-a)(x-b)}{(c-a)(c-b)}.c+ \dfrac {(x-b)(x-c)}{(a-b)(a-c)}.a+\dfrac {(x-a)(x-c)}{(b-a)(b-c)}.b=x$$
My Attempt:
Proving $ \dfrac {(x-a)(x-b)}{(c-a)(c-b)}.c+ \dfrac {(x-b)(x-c)}{(a-b)(a-c)}.a+\dfrac {(x-a)(x-c)}{(b-a)(b-c)}.b=x$ is equivalent to proving
$$ \dfrac {(x-a)(x-b)}{(c-a)(c-b)}.c+ \dfrac {(x-b)(x-c)}{(a-b)(a-c)}.a+\dfrac {(x-a)(x-c)}{(b-a)(b-c)}.b-x=0$$
i.e $f(x)=0$
where,
$$f(x)= \dfrac {(x-a)(x-b)}{(c-a)(c-b)}.c+ \dfrac {(x-b)(x-c)}{(a-b)(a-c)}.a+\dfrac {(x-a)(x-c)}{(b-a)(b-c)}.b-x$$
..

Let $f(x) = \dfrac {(x-a)(x-b)}{(c-a)(c-b)}.c+ \dfrac {(x-b)(x-c)}{(a-b)(a-c)}.a+\dfrac {(x-a)(x-c)}{(b-a)(b-c)}.b-x = 0$
Note that $f(x)$ seems to be a  quadratic equation in $x$ and  can have no more than two roots. But $f(a) = f(b) = f(c) = 0$ (as Lord Shark stated in comment), meaning that $a,b,c$ are three roots. Since a quadratic can not have more than two roots, this is an identity in $x$, valid for all $x \in R$.
