Let $a,$ $b,$ $c$ be real numbers such that

Question

$$\begin{array}{rl}{a}^2-bc& =42,\\ b^2-ac& =-5,\\ c^2-ab& =25.\end{array}$$
Find $a^2 + b^2 + c^2.$

Answer 1

If we sum the equations, we get $(a^2 - bc) + (b^2 - ac) + (c^2 - ab) = 42 - 5 + 25,$ so
$$a^2+b^2+c^2-(ab+ac+bc)=62.$$Now we can double each equation to get
$$\begin{array}{rl}2a^2-2bc& =84,\\ 2b^2-2ac& =-10,\\ 2c^2-2ab& =50.\end{array}$$
Adding these equations, we get $(2a^2 - 2bc) + (2b^2 - 2ac) + (2c^2 - 2ab) = 84 - 10 + 50,$ so $2(a^2 + b^2 + c^2) - 2(ab + ac + bc) = 124.$ Then $a^2 + b^2 + c^2 - (ab + ac + bc) = 62.$ Hence,
$$(a-b{)}^2+(b-c{)}^2+(c-a{)}^2=0.$$This forces $a = b = c.$ Substituting into one of the original equations, we get $a^2 - a^2 = 42,$ so $a^2 = 21.$ Hence,
$$a^2+b^2+c^2=\boxed{{\displaystyle 63}}.$$