If $ a + b + c = 0$, then $(\frac{2a^2}{3bc}+\frac{2b^2}{3ca}+\frac{2c^2}{3ab})$ is equal to :

3 4 1 2

2

If $ a + b + c = 0$, then $(\frac{2a^2}{3bc}+\frac{2b^2}{3ca}+\frac{2c^2}{3ab})$ is equal to =? If a + b + c = 0 Then, a3 + b3 + c3 = 3abc $(\frac{2a^2}{3bc}+\frac{2b^2}{3ca}+\frac{2c^2}{3ab})$ = \(\frac{2a^3 + 2b^3 + 2c^3}{3abc}\) = 2×\(\frac{3abc}{3abc}\) = 2