Question

Answer 1

`-> a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 3ab + 3bc + 3ca`

`= a^2 + b^2 + c^2 - ab - bc - ca = 0`

`-> 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca = 0`

`-> (a-b)^2 + (b-c)^2 + (c-a)^2 = 0`

`-> {(a-b=0), (b-c=0), (c-a = 0):}`

`<=> a = b = c`

Answer 2

(a+b+c)^2=3(ab+bc+ca)`

`<=>a^2+b^2+c^2+2(ab+bc+ca)=3(ab+bc+ca)`

`<=>a^2+b^2+c^2-(ab+bc+ca)=0`

`<=>a^2+b^2+c^2-ab-bc-ca=0`

`<=>2a^2+2b^2+2c^2-2ab-2bc-2ca=0`

`<=>(a-b)^2+(b-c)^2+(a-c)^2=0`

`=>a=b=c`

Answer 3

\(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)

=>a^2+b^2+c^2-ab-bc-ac=0

=>2a^2+2b^2+2c^2-2ab-2bc-2ac=0

=>(a-b)^2+(b-c)^2+(a-c)^2=0

=>a=b=c