Question

Prove If (a-b)^2 +(b-c)^2 +(c-a)^2=0
Then a=b=c.

Answer 1

Step-by-step explanation:

$(a - b) {}^{2} + (b - c) {}^{2} + (c - a) { }^{2} = 0$

$a {}^{2} + b {}^{2} - 2ab + b {}^{2} + c {}^{2} - 2bc + c {}^{2} + a {}^{2} - 2ca = 0$

$2(a {}^{2} + b {}^{2} + c {}^{2}) = 2(ab + bc + ca)$

$a {}^{2} + b {}^{2} + c {}^{2} = ab + bc + ca$

Compairing both sides....

$a {}^{2} = ab \\ = > a = b \\ b {}^{2} = bc \\ = > b = c \\ a = b = c$

Proved