Question

(x+y-z) ^2+(y+z-x)^2+(z+x-y)^2=0 then find x+y+z

Answer 1

Please see the attachment

Answer 2

/* We know that */

$\boxed { \pink { If a^{2} + b^{2} + c^{2} = 0 \implies a = b = c = 0 }}$

$Here , a = x + y - z , \: b = y + z - x \\and \: z + x - y = c$

$(x+y-z) ^2+(y+z-x)^2+(z+x-y)^2=0 \:(given)$

$\implies x + y - z = 0 \: --(1)\: , \: y + z - x = 0 \: --(2) \\and \: z + x - y = 0 \: ---(3)$

/* Add Equations (1) , (2) and (3) , we get */

$\implies x + y - z + y + z - x + z + x - y = 0$

$\implies x + y + z = 0$

$Hence \: proved$

•••♪