I understand that this is too easy a question for most of us here but still I ask it for one reason.
I thought that putting $x=a,x=b,x=c$ makes the above equation $=0$ in each case. Hence number of distinct, real $x$ for which the equation is $=0$ should be $=3$ but the answer is $>3$.I am a little confused why? Any light on this small query?Or to put it straight where am I wrong?
I think I got the answer , yes for sure putting $x=a,x=b,x=c$ makes the equation $=0$ in each case. But fundamentally it can also happen that for some real $x=x_0 \ne a \space or \space b \space or \space c$ for which $\frac{(x-a)(x-b)}{(c-a)(c-b)}+\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}=1$.Hence that makes the number of distinct real $x$ for which the equation is 0