$$z_1$$ and $$z_2$$ are the roots of $${3z^2} +3z + b = 0.$$    If $$O\left( 0 \right),A\left( {{z_1}} \right),B\left( {{z_2}} \right)$$    form an equilateral triangle, then the value of $$b$$ is

Solution :
$$\eqalign{ & {z_1} + {z_2} = - 1,{z_1}{z_2} = \frac{b}{3} \cr & {0^2} + {z_1}^2 + {z_2}^2 = 0 \times {z_1} + 0 \times {z_2} + {z_1}{z_2} \cr & \Rightarrow {\left( {{z_1} + {z_2}} \right)^2} - 2{z_1}{z_2} = {z_1}{z_2} \cr & \Rightarrow 1 = 3{z_1}{z_2} = 3\frac{b}{3} \cr & \Rightarrow b = 1. \cr} $$