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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

1 Answer

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by (23.5k points)
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Best 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.

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