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{align*}
2a^2 - 2bc &= 84, \\
2b^2 - 2ac &= -10, \\
2c^2 - 2ab &= 50.
\end{align*}
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,
\[a^2 + b^2 + c^2 = \boxed{63}.\]
Let $a,$ $b,$ $c$ be real numbers such that
\begin{align*}
a^2 - bc &= 42, \\
b^2 - ac &= -5, \\
c^2 - ab &= 25.
\end{align*}
Find $a^2 + b^2 + c^2.$
1 answer
1 answer