Question:
If
\[ a^2+b^2+c^2-ab-bc-ca=0, \] then
(a) \[ a+b=c \]
(b) \[ b+c=a \]
(c) \[ c+a=b \]
(d) \[ a=b=c \]
Solution:
Using identity:
\[ a^2+b^2+c^2-ab-bc-ca = \frac{1}{2} \left[ (a-b)^2+(b-c)^2+(c-a)^2 \right] \]
Given:
\[ a^2+b^2+c^2-ab-bc-ca=0 \]
Therefore,
\[ \frac{1}{2} \left[ (a-b)^2+(b-c)^2+(c-a)^2 \right] =0 \]
Since sum of squares can be zero only when each term is zero,
\[ a-b=0 \]
\[ b-c=0 \]
\[ c-a=0 \]
Hence,
\[ a=b=c \]
Therefore, the correct answer is:
\[ \boxed{a=b=c} \]