Question : If $\frac{x}{(b–c)(b+c–2a)}=\frac{y}{(c–a)(c+a–2b)}=\frac{z}{(a–b)(a+b–2c)}$, then $(x+y+z)$ is:
Option 1: $a+b+c$
Option 2: $0$
Option 3: $a^2+b^2+c^2$
Option 4: $2$
Correct Answer: $0$
Solution :
Given: $\frac{x}{(b–c)(b+c–2a)}=\frac{y}{(c–a)(c+a–2b)}=\frac{z}{(a–b)(a+b–2c)}$
Let $\frac{x}{(b–c)(b+c–2a)}=\frac{y}{(c–a)(c+a–2b)}=\frac{z}{(a–b)(a+b–2c)}=k$.
${x}=k{(b–c)(b+c–2a)}$ (equation 1)
${y}=k{(c–a)(c+a–2b)}$ (equation 2)
${z}=k{(a–b)(a+b–2c)}$ (equation 3)
Adding equations 1, 2, and 3 respectively, we get,
$x+y+z=k{(b–c)(b+c–2a)}+k{(c–a)(c+a–2b)}+k{(a–b)(a+b–2c)}$
$x+y+z=k[(b–c)(b+c–2a)+(c–a)(c+a–2b)+(a–b)(a+b–2c)]$
$x+y+z= k[b^2+bc–2ab–bc–c^2+2ac+c^2+ac–2bc–ac–a^2+2ab+a^2+ab–2ac–ab–b^2+2bc)$
$x+y+z$ = 0
Hence, the correct answer is 0.
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