If \(a-b=4\), \(b-c=6\) and \(a-c=10\), then the value of \(a^{2}+b^{2}+c^{2}-ab-bc-ca\) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 76
The expression can be written with the three given differences through the identity
\(\displaystyle a^{2}+b^{2}+c^{2}-ab-bc-ca=\frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]\)
(opening the three squares gives each of \(a^{2}, b^{2}, c^{2}\) twice and each product \(-2ab, -2bc, -2ca\) once; halving gives the expression).
\((a-b)^{2}=4^{2}=16\)
\((b-c)^{2}=6^{2}=36\)
\((c-a)^{2}=(a-c)^{2}=10^{2}=100\)
Value \(\displaystyle =\frac{1}{2}(16+36+100)=\frac{152}{2}=76\)