Solve $3x^3-26x^2+52x-24=0$the roots are G.P.

Given $3x^3-26x^2+52x-24=0$

Given that the roots are in G.P.

$\frac{a}{r}, a, ar$are the roots of the equation 

$S_3=\frac{a}{r}.a.ar=\frac{-(-24)}{3}{S}_3=\frac{-d}{a}$

$a^3=8\Rightarrow a^3=2^3\Rightarrow a=2$

$(x-2)(3x^2-20x+12)=0$

$\begin{array}{l}(x-2)(3x^2-18x-2x+12)=0\\ (x-2)(x-6)(3x-2)=0\end{array}$

The roots are $x=2, 6, 1/3$