Let the GP be $a,ar,ar^2,\ldots,$ where $0 < r<1.$ Then, $a + ar + ar^2 + \ldots = 3$ and $a^2 + a^2r^2+a^2r^4 + \ldots = 9/2.$
$\quad \implies \dfrac{a}{1-r} = 3 \qquad$ and $\qquad\dfrac{a^2}{1-r^2} = \dfrac{9}{2}$
$\quad \implies \dfrac{9(1-r)^2}{1-r^2} = \dfrac{9}{2} \implies \dfrac{1-r}{1+r} = \dfrac{1}{2} \implies r = \dfrac{1}{3}.$
Putting $r = \dfrac{1}{3}$ in $\dfrac{a}{1-r} = 3,$ we get $a = 2.$
Now the required sum of the cubes is $a^3 + a^3r^3+a^3r^6 + \ldots = \dfrac{a^3}{1-r^3} = \dfrac{8}{1-\frac{1}{27}}=\dfrac{108}{13}$
