[tex]\sf a + b + c = 0\\abc \neq 0\\\\ \frac{a^2}{bc} + \frac{b^2}{ac} + \frac{c^2}{ab} = \frac{a^3+b^3+c^3}{abc} \\\\\sf Formula: \\a^3 + b^3 + c^3 = (a + b + c) (a^2 + b^2 + c^2- ab - bc - ca) + 3abc\\\\\frac{a^3+b^3+c^3}{abc}= \frac{(a + b + c) (a^2 + b^2 + c^2- ab - bc - ca) + 3abc}{abc}\\\\\sf However, ~a+b+c = 0. ~So:\\\\\frac{0 (a^2 + b^2 + c^2- ab - bc - ca) + 3abc}{abc}\\\\\sf Multiply~ anything ~by~ 0, ~and~ you ~get~ 0. ~So~ simplifying:\\\\\frac{3abc}{abc} = \boxed{3}[/tex]
Your final answer is 3.
