Solve the roots of a cubic polynomial? I have had trouble with this question - mainly due to the fact that I do not fully understand what a 'geometric progression' is:
"Solve the equation $x^3 - 14x^2 + 56x - 64 = 0$" if the roots are in geometric progression.
Any help would be appreciated.
 A: It means the roots are of the form $a, ar, ar^2$.
Here it is not too difficult to see that $2, 4, 8$ are ok, just look at the constant term, which is $- (a \cdot ar \cdot ar^2) = - (a r)^3$, and check that $2, 4, 8$ fit with the other coefficients
$$
14 = 2 + 4 + 8,
\qquad
56 = 2 \cdot 4 + 2 \cdot 8 + 4 \cdot 8.
$$
A: Let the roots be $a,a\cdot r,a\cdot r^2$
Using Vieta's formula  $a+a\cdot r+a\cdot r^2=14\implies a(1+r+r^2)=14$
and $a(a\cdot r)+a\cdot r(a\cdot r^2)+a(a\cdot r^2)=56\implies a^2\cdot r(1+r+r^2)=56$
On division, $ar=4$ as $a\cdot r\ne0$
$$\implies a=\frac 4r\implies \frac{4(1+r+r^2)}r=14\implies 2r^2-5r+2=0\implies r=2\text{  or }\frac12$$

Alternatively, 
 using Vieta's formula $a\cdot(a\cdot r) \cdot (a\cdot r^2)=64\implies (ar)^3=64$
So, $a\cdot r$ can be one of $4,4w,4w^2$ where $w$ is the cube root of $1$
Using Polynomial Remainder Theorem, observe that $4$ is a root of the given equation
$\implies a\cdot r=4\iff a=\frac4r$
Again, using Vieta's formula $a+a\cdot r+a\cdot r^2=14\implies 2r^2-5r+2=0$
