If \(a=1044\), \(b=1047\) and \(c=1041\), then find the value of \(a^2+b^2+c^2-ab-bc-ca\).
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 27
The expression depends only on the differences of the numbers: \(a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]\). The differences here are small, so no large squares are needed.
\(a-b=1044-1047=-3\), \(b-c=1047-1041=6\), \(c-a=1041-1044=-3\)
\((a-b)^2+(b-c)^2+(c-a)^2=9+36+9=54\)
Required value \(\displaystyle=\frac{54}{2}=27\)