(c) 0
Let \(\frac{x}{(b-c)(b+c-2a)}=\frac{y}{(c-a)(c+a-2b)}=\frac{z}{(a-b)(a+b-2c)}\) = k.
Then, x = k(b – c) (b + c – 2a)
y = k(c – a) (c + a – 2b)
z = k(a – b) (a + b – 2c)
∴ x + y + z = k(b – c) (b + c – 2a) + k(c – a) (c + a – 2b) + k(a – b) (a + b – 2c)
= k(b2 – c2 – 2ab + 2ca) + k(c2 – a2 – 2bc + 2ab) + k (a2 – b2 – 2ca + 2bc)
= k(b2 – c2 – 2ab + 2ca + c2 – a2 – 2bc + 2ab + a2 – b2 – 2ca + 2bc)
= k × 0 = 0.