If $a^{2} + b^{2} + c^{2} - a b - b c - c a = 0$ , prove that $a = b = c$ .

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Solution

Consider, $a^{2} + b^{2} + c^{2} - a b - b c - c a = 0$
Multiply both sides with $2$ , we get
$2 (a^{2} + b^{2} + c^{2} - a b - b c - c a) = 0$
$\Rightarrow$ $2 a^{2} + 2 b^{2} + 2 c^{2} - 2 a b - 2 b c - 2 c a = 0$
$\Rightarrow$ $(a^{2} - 2 a b + b^{2}) + (b^{2} - 2 b c + c^{2}) + (c^{2} - 2 c a + a^{2}) = 0$
$\Rightarrow$ $(a - b)^{2} + (b - c)^{2} + (c - a)^{2} = 0$
Since the sum of square is zero then each term should be zero
$\Rightarrow$ $(a - b)^{2} = 0, (b - c)^{2} = 0, (c - a)^{2} = 0$
$\Rightarrow$ $(a - b) = 0, (b - c) = 0, (c - a) = 0$
$\Rightarrow$ $a = b, b = c, c = a$
$∴$ $a = b = c .$

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