Step 1: Write the GP roots smartly.
The three roots are in geometric progression, so write them as $\dfrac{a}{r},\ a,\ ar.$ This makes their product simple.
Step 2: Recall the root rules for a cubic.
For $x^3-px^2+qx-s=0$: sum of roots $=p$, sum of products in pairs $=q$, product of all $=s$.
Step 3: Product of roots.
$\dfrac{a}{r}\cdot a\cdot ar=a^3.$ So $s=a^3$, which means $a=s^{1/3}.$
Step 4: Sum of roots.
$p=\dfrac{a}{r}+a+ar=a\left(\dfrac1r+1+r\right).$
Step 5: Sum of products in pairs.
$q=\dfrac{a}{r}\cdot a+a\cdot ar+\dfrac{a}{r}\cdot ar=a^2\left(\dfrac1r+r+1\right).$ The same bracket appears, so $q=a\cdot p.$
Step 6: Remove $a$.
From $q=ap$ cube both sides: $q^3=a^3p^3.$ Since $a^3=s$, we get \[ q^3=p^3 s. \] \[ \boxed{q^3=p^3 s} \]