I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.
If $a, b, c$ are distinct numbers such that $a^2 - bc = 2014$, $b^2 + ac = 2014$, $c^2 + ab = 2014$. Then compute $a^2 + b^2 + c^2$
(A)$4030$ (B)$4028$ (C)$4026$ (D)$4000$ (E)$2014$
Adding these three equations together
$$a^2 + b^2 + c^2 + ab + ac - bc = 3\times2014 \quad(1)$$
And also found that
\begin{align} (a-b-c)^2 &= a^2 + b^2 + c^2 - 2ab - 2ac + 2bc\\ (a-b-c)^2 &= a^2 + b^2 + c^2 - 2(ab + ac - bc)\\ \end{align}
I don't know how to continue to reduce $ ab + ac - bc$, or am I using the wrong way to reducing it?
I'm very appreciate for those who have helped me to hint/explain me on how to do all these questions (I'm currently 10th grade (in US grade system), so I don't understand these much, all of these are outside my syllabus)
Note that $$b^2+ac=c^2+ab\iff b^2-c^2+ac-ab=0\iff (b-c)(b+c)-a(b-c)=0$$ $$\iff (b-c)(b+c-a)=0\iff a=b+c.$$
Now $$a^2+b^2+c^2=(b+c)^2+b^2+c^2=2(b^2+c^2+bc)$$ $$=2((b+c)^2-bc)=2(a^2-bc)=2\times 2014$$