Question : If $a+b+c=6$ and $a^2+b^2+c^2=14$, then what is the value of $(a-b)^2+(b-c)^2+(c-$ a) ${ }^2$?
Option 1: –8
Option 2: 8
Option 3: 10
Option 4: 6
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Correct Answer: 6
Solution : $a^2+b^2+c^2=14$ $a+b+c=6$ Squaring on both sides, $(a+b+c)^2=6^2$ ⇒ $a^2+b^2+c^2+2ab+2bc+2ca = 36$ ⇒ $14 + 2ab+2bc+2ca = 36$ ⇒ $2ab+2bc+2ca = 36-14 = 22$ --------------------(i) $(a-b)^2+(b-c)^2+(c-$ a) ${ }^2$ $= a^2+b^2-2ab+b^2+c^2-2bc+c^2+a^2-2ca$ $= 2(a^2+b^2+c^2)-(2ab+2bc+2ca)$ $=2×14 – 22$ $= 6$ Hence, the correct answer is 6.
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