If $\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma }$ are the roots of ${\mathrm{x}}^3 + 2{\mathrm{x}}^2 – 4\mathrm{x} –3= 0$ then the equation whose roots are $\frac{\mathrm{\alpha }}{3},\frac{\mathrm{\beta }}{3},\frac{\mathrm{\gamma }}{3}$ is

$\begin{array}{l}Let\alpha ,\beta ,\gamma arerootsoff\left(x\right)=x^3+2x^2-4x-3=0\\ Thentheequationwhoserootsare\frac{\alpha }{3},\frac{\beta }{3},\frac{\gamma }{3}isf\left(3x\right)=0\\ \Rightarrow {\left(3x\right)}^3+2{\left(3x\right)}^2-4\left(3x\right)-3=0\\ \Rightarrow 27x^3+18x^2-12x-3=0\\ \Rightarrow 9x^3+6x^2-4x-1=0\end{array}$