Dao Trong Luan, you missed the memorable identity : \(\left(b-c\right)^2=\left(b-c\right)\left(b+c\right)\) ?

My solution : \(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)

\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=0\)

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

Moreover : \(\left(b-c\right)^2+\left(c-a\right)^2+\left(a-b\right)^2\)

\(=b^2-2bc+c^2+c^2-2ac+a^2+a^2-2ab+b^2\)

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

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

So, the answer is \(\dfrac{1}{3}\)

Dao Trong Luan, you missed the memorable identity : (b−c)2=(b−c)(b+c)

 ?

My solution : a+b+c=0⇒(a+b+c)2=0

⇒a2+b2+c2+2ab+2bc+2ca=0

⇒a2+b2+c2=−2(ab+bc+ca)

Moreover : (b−c)2+(c−a)2+(a−b)2

=b2−2bc+c2+c2−2ac+a2+a2−2ab+b2

=2(a2+b2+c2)−2(ab+bc+ca)

=2(a2+b2+c2)+(a2+b2+c2)=3(a2+b2+c2)

So, the answer is 13

a+b+c = 0

\(\Rightarrow\left(b-c\right)^2+\left(c-a\right)^2+\left(a-b\right)^2\)

\(=\left(b-c\right)\left(b+c\right)+\left(c-a\right)\left(c+a\right)+\left(a-b\right)\left(a+b\right)\)

\(=\left(b-c\right)\left(0-a\right)+\left(c-a\right)\left(0-b\right)+\left(a-b\right)\left(0-c\right)\)

\(=-a\left(b-c\right)-b\left(c-a\right)-c\left(a-b\right)\)

\(=-ab+ac-bc+ab-ac+bc\)

\(=\left(-ab+ab\right)+\left(-bc+bc\right)+\left(-ac+ac\right)=0+0+0=0\)

\(\Rightarrow Q=\dfrac{\left(a^2+b^2+c^2\right)}{0}=????\)

Unsatisfactory, so Q is void