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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