If \(a=618\), \(b=616\) and \(c=617\), then find the value of \(4(a^2+b^2+c^2-ab-bc-ca)\).
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 12
The expression in the bracket can be written with the differences alone: \(a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]\), so the large numbers never need to be squared.
\(a-b=2\), \(b-c=-1\), \(c-a=1\)
\(a^2+b^2+c^2-ab-bc-ca=\displaystyle\frac{1}{2}(4+1+1)=3\)
\(4\times3=12\)