Question

please solve it fast​

Answer 1

Answer:

PROVED BELOW

Step-by-step explanation:

$a^{2}+b^{2}+c^{2}-ab-bc-ca=0\\=> a^{2}-ab+b^{2}-bc+c^{2}-ca=0\\=>a(a-b)+b(b-c)+c(c-a)=0\\=>a-b=0(\;because\;the\;equation\;is\;equal\;to\;0\\so,\';each\;no.\;should\;be\;zero\;)\\\;\;\;=>a=b...................(i)\\also\\b-c=0\;\;=>b=c..............(ii)\\\also\\c-a=0\\=>c=a.............................(iii)\\From\;(i)\;(ii)\;and(iii)\\a=b=c\\HENCE \;\;PROVED!!!$

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