If a + b - c = 0, then what is the value of $\frac{(b-c)^2}{4bc}+\frac{(c-a)^2}{4ca}-\frac{(a+b)^2}{4ab}$ ?

$-\frac{3}{4}$

If a + b - c = 0, then what is the value of $\frac{(b-c)^2}{4bc}+\frac{(c-a)^2}{4ca}-\frac{(a+b)^2}{4ab}$

= If a + b = c

Put the value of a = 1, b = 1 and c = 2 ( These value will satisfy the given equation )

$\frac{(2-1)^2}{4(1)(2)}+\frac{(2-1)^2}{4(2)(1)}-\frac{(1+1)^2}{4(1)(1)}$

= \(\frac{1}{8}\) + \(\frac{1}{8}\) - 1

= \(\frac{1}{4}\) - 1 = $-\frac{3}{4}$