Simplify this easy expression. It came in NSTSE 2015 Class 9 WB

Since the given expression is true for all numbers ,

Let $a = 2 , b = 1 , c = 0$ .

Hence the given expression becomes :

$\dfrac{(2+1)^2}{(1-0)(0-2)} + \dfrac{(1+0)^2}{(2-1)(0-2)} + \dfrac{(0+2)^2}{(2-1)(1-0)}$

$= \dfrac{(3)^2}{(1)(-2)} + \dfrac{(1)^2}{(1)(-2)} + \dfrac{(2)^2}{(1)(1)}$

$= -\dfrac{9}{2} - \dfrac{1}{2} + \dfrac{4}{1}$

$= -5 + 4 = \huge\boxed{-1}$

Just plug in a=1, b=2, c=3 and evaluate.

The numerator will be f(a,b,c)=(a+b)²(a-b)+(b+c)²(b-c)+(c+a)²(c-a). This function is a quadratic polynomial in a, b or c variable (note that each quadratic term will be cancelled). Doing b=a, f(a,a,c)=(a+c)²(a-c)+(c+a)²(c-a)=0. Doing c=a, f(a,b,a)=0 and finally doing c=b, f(a,b,b)=0. So f(a,b,c)=k(a-b)(b-c)(c-a). Take f(-1,0,1)=-1-1=-2=k(-1)(-1)2 <--> k=-1 so f(a,b,c)=(-1)(a-b)(b-c)(c-a) and the whole expression will be reduced to -1.

Answer : -1. Solution : (a+b)^2 / (b-c)(c-a) + (b+c)^2 / (a-b)(c-a) + (c+a)^2 / (a-b)(b-c): = (a+b)^2 (a-b) + (b+c)^2 (b-c) + (c+a)^2*(c-a) / (a-b)(b-c)(c-a). -> (a+b)(a+b)(a-b) + (b+c)(b+c)(b-c) + (c+a)(c+a)(c-a) / (a-b)(b-c)(c-a). -> (a+b)(a^2-b^2) + (b+c)(b^2-c^2) + (c+a)(c^2-a^2) / (a-b)(b-c)(c-a). -> a^3-ab^2+a^2b-b^3 + b^3-bc^2+b^2c-c^3 + c^3-ca^2+c^2a-a^3 / (a-b)(b-c)(c-a) -> -ab^2+a^2b-bc^2+b^2c-ca^2+c^2a / (a-b)(b-c)(c-a). -> (b-a)(b-c)(c-a) / (a-b)(b-c)(c-a). -> (b-a) / (a-b) = -1.