If \(\frac{x}{(b-c)(b+c-2a)}=\frac{y}{(c-a)(c+a-2b)}=\frac{z}{(a-b)(a

1.	If \(\frac{x}{(b-c)(b+c-2a)}=\frac{y}{(c-a)(c+a-2b)}=\frac{z}{(a-b)(a+b-2c)}\) , what is the value of x + y + z ?(a) (a + b + c) (b) a2 + b2 + c2 (c) 0 (d) 1
Answer» (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.

If \(\frac{x}{(b-c)(b+c-2a)}=\frac{y}{(c-a)(c+a-2b)}=\frac{z}{(a-b)(a+b-2c)}\) , what is the value of x + y + z ?(a) (a + b + c) (b) a2 + b2 + c2 (c) 0 (d) 1

Answer»

(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.