If a, $b$ and $c$ are non-zero numbers with $a+b+c=0$, then $a^2/(bc)+b^2/(ca)+c^2/(ab)$ equals:

Put the three fractions over the common denominator $abc$: the sum becomes $(a^3+b^3+c^3)/(abc)$. When $a+b+c=0$ the identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ forces $a^3+b^3+c^3=3abc$, so the value is $3abc/abc=3$. (0 comes from expecting the cubes to vanish along with the sum; −3 flips a sign that never appears.) Test a, $b,c=1,1,-2:1/(-2)+1/(-2)+4/1=-1+4=3$.