Question

if a^2 +b^2+c^2 -ab-bc-ca=0,Provethat a=b=c

Answer 1

${a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca = 0 \\ \\ Multiplying \: whole \: expression \: by \: 2 \: , \\ \\ 2 {a}^{2} + 2{b}^{2} + 2 {c}^{2} -2 ab - 2bc -2 ca = 0 \\ \\ {a}^{2} - 2ab + {b}^{2} + {b}^{2} - 2bc + {c}^{2} + {c}^{2} - 2ca + {a}^{2} = 0 \\ \\ {( \: a - b \: )}^{2} + {( \: b - c \: )}^{2} + {( \: c - a \: )}^{2} = 0 \\ \\ Square \: of \: a \: number \: is \: always \: \\ greater \: than \: or \: equa l \: to \: 0 \\ \\ Therefore \: , \: all \: above \: terms \: are \: \\ equal \: to \: 0 \\ \\ a - b = 0\: \\ b - c \: = 0 \\ c - a = 0 \\ \\ \large \boxed{Therefore \: \: , \: \: a = b = c = 0} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: Proved$