Question : If $(a+b+c)=19$ and $\left(a^2+b^2+c^2\right)=155$, find the value of $(a-b)^2+(b-c)^2+(c-a)^2$
Option 1: 104
Option 2: 108
Option 3: 100
Option 4: 98
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Correct Answer: 104
Solution :
Given: $(a+b+c)=19$
$(a^2+b^2+c^2)=155$
We know that $(a+b+c)^2 = (a^2+b^2+c^2)+2(ab+bc+ca)$
So, $19^2 = 155+2(ab+bc+ca)$
⇒ $ab+bc+ca = \frac{361-155}{2}$
⇒ $ab+bc+ca= 103$
Now, $(a-b)^2+(b-c)^2+(c-a)^2$
$= a^2+b^2-2ab+b^2+c^2-2bc+c^2+a^2-2ca$
$= 2(a^2+b^2+c^2)-2(ab+bc+ca)$
$= 2(155)-2(103)$
$= 2 × 52$
$= 104$
Hence, the correct answer is 104.
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