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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