2025/032) How do I find the roots of the equation $x^3-14x^2+56x–64=0$, the roots are in geometric progression?

 roots are in GP so sw have root are $r, \frac{r}{a},ra$ so we get product of roots 

=$r^3= 64$ or $r=4$ So one root is 4 and hence (x-4) is a factor.  

we have by synthetic division.

$x^3-14x^2+56x–64=0= (x-4)(x^2-10x + 16)$

one factor is (x-4) and other factor is quadratic  and can be factored as $(^2-10x+16=(x-2)(x-8)$

giving roots 2,4,8 and they are in GP(just for cross checking)