\(\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} = 3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\times abc \\ \\
\frac{a^2 \times abc}{bc} + \frac{b^2 \times abc}{ca} + \frac{c^2 \times abc}{ab}=3abc \\ \\
a^2 \times a + b^2 \times b + c^2 \times c = 3abc \\ \\
a^3+b^3+c^3=3abc \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |-3abc \\ \\
a^3+b^3+c^3-3abc=0 \\ \\\)
You know the value of a+b+c, so try to factor a+b+c out of the expression:
\(a^3+b^3+c^3-3abc= \\ \\ a^3+a^2b+a^2c+ab^2+b^3+b^2c+ac^2+bc^2+c^3-a^2b-ab^2-abc \\ -abc-b^2c-bc^2-a^2c-abc-ac^2= \\ \\
a^2(a+b+c)+b^2(a+b+c)+c^2(a+b+c)-ab(a+b+c) \\
-bc(a+b+c)-ac(a+b+c)= \\ \\
(a^2+b^2+c^2-ab-bc-ac)(a+b+c)\)
Back to the equation:
\((a^2+b^2+c^2-ab-bc-ac)(a+b+c)=0\)
You know that a+b+c=0.
\((a^2+b^2+c^2-ab-bc-ac) \times 0=0 \\ \\
0=0 \\ \\
LHS=RHS\)
The equation \(\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} =3\) is true for any values of a, b, c, such that a+b+c=0. Proved.
