For real a, b and c if $$a^2  + b^2 + c^2$$ = ab + bc + ca, then find the value of $$(a + b + c)^2$$

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

Multiply both sides with 2, we get

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

$$(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)=0$$

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

Since the sum of square is zero then each term should be zero

$$(a-b)^2=0$$

$$(b-c)^2=0$$

$$(c-a)^2=0$$

$$a=b=c$$--------1

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

from equation 1

$$=a^2+a^2+a^2+2\times3a^2$$

$$=9a^2$$