Question

If x + y + z = 0, Prove that ((x + y)^2)/xy + ((y+z)^2)/yz + ((z+x)^2)/zx = 3

Answer 1

$x + y + z = 0 \\ x + y = - z \\ z + y = - x \\ x + z = - y \\ \frac{ {(x + y)}^{2} }{xy} = \frac{ {( - z)}^{2} }{xy} = {z \div xy}^{2} \\ similarly \:we\: can \:find \:other \:values\\so \: required \: answer \: is \\ \frac{ {z}^{2} }{xy} + \frac{ {x}^{2} }{yz} + \frac{ {y}^{2} }{xz} = \frac{( {x}^{3} + {y}^{3} + {z}^{3}) }{xyz} \\ = \frac{3xyz}{xyz} = 3 \\ since \: if \: x + y + z = 0 \: then \: {x}^{3} + {y}^{3} + {z}^{3} \\= 3xyz$
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