This can be interesting too

\hspace{-16}$Using Repeated Root Theorem......\\\\ If $\bf{x=\alpha}$ is a Root of $\bf{p(x)=0}$. Then\\\\ $\bf{x=\alpha}$ is also a Root of $\bf{p\hspace{-10}\quad'(x)=0}$......\\\\ So Here $\bf{x=\alpha}$ is a Root of $\bf{p(x)=x^3+3ax^2+3bx+c=0}$\\\\ So $\bf{x=\alpha}$ is also a Root of $\bf{.p\hspace{-10}\quad'(x)=3x^2+6ax+3b=0}$\\\\ So $\bf{x=\alpha}$ is a Root of $\bf{x^2+2ax+b=0...................(1)}$\\\\ So option $\bf{(a)}$ is Right.......\\\\ Similarly for $\bf{(b)}$ and $\bf{(c)}$\\\\ If $\bf{x=\alpha}$ is a Root of $\bf{x^3+3ax^2+3bx+c=0}$, Then\\\\ $\bf{\alpha^3+3a\alpha^2+3b\alpha+c=0......................(2)}$\\\\ and Multiply $\bf{(1)}$ by $\bf{\alpha}$, We Get\\\\ $\bf{\alpha^3+2a\alpha^2+b\alpha = 0........................(3)}$\\\\ Sub $\bf{(1)-(2)....}$\\\\ $\bf{a\alpha^2+2b\alpha+c=0............................(4)}$\\\\ So $\bf{x=\alpha}$ is also a Root of $\bf{ax^2+2bx+c=0}$\\\\

\hspace{-16}$Now $\bf{\alpha^2+2a\alpha+b=0.....................}$ from $\bf{(1)}$\\\\ and $\bf{\alpha^2+\frac{2b}{a}\alpha+\frac{c}{a}=0.....................}$ from $\bf{(4)}$\\\\ Substract These $\bf{2}$ Equations.........\\\\ $\bf{\alpha=\frac{1}{2}.\left(\frac{c-ab}{a^2-b}\right)}$\\\\ So options $\bf{(a)\;\;(b)}$ and $\bf{(d)}$ are Right.............