Question:
If \[ (a-b)^2+(b-c)^2+(c-a)^2=0 \] then ___________
Solution:
We know that the square of any real number is always non-negative.
Therefore,
\[ (a-b)^2 \ge 0 \]
\[ (b-c)^2 \ge 0 \]
\[ (c-a)^2 \ge 0 \]
Their sum is given as:
\[ (a-b)^2+(b-c)^2+(c-a)^2=0 \]
This is possible only when each term is zero.
So,
\[ a-b=0 \]
\[ b-c=0 \]
\[ c-a=0 \]
Hence,
\[ a=b=c \]