Question : If $x+y+z = 22$ and $xy+yz+zx = 35$, then what is the value of $\small (x-y)^{2}+(y-z)^{2}+(z-x)^{2}$?
Option 1: 793
Option 2: 681
Option 3: 758
Option 4: 715
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Correct Answer: 758
Solution :
Given: $x+y+z = 22$ and $xy+yz+zx = 35$
We know that $(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx)$
⇒ $22^2 = x^2+y^2+z^2+2\times 35$
⇒ $484 = x^2+y^2+z^2+70$
⇒ $x^2+y^2+z^2 = 484-70 = 414$
So, $(x-y)^2+(y-z)^2+(z-x)^2 = 2[x^2+y^2+z^2-(xy+yz+zx)]$
⇒ $(x-y)^2+(y-z)^2+(z-x)^2 = 2[414-35]$
$\therefore(x-y)^2+(y-z)^2+(z-x)^2 = 758$
Hence, the correct answer is 758.
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