Find roots of 3 degree polynomial such that they are in geometric progression I have the polynomial $P(x) = x^3 + mx^2-3x+1, m\in\mathbb{R}$. I need to find $m$ such that the roots of that polynomial are in geometric progression.
My attemp of solving this was to use Vieta's formulas. So  $x_1+x_2+x_3 = -m, x_1x_2+x_1x_3+x_2x_3 = -3, x_1x_2x_3 = -1$. If $x_1, x_2, x_3$ are in geometric progression then let $x_1 = \alpha, x_2 = q\alpha, x_3 = q^2\alpha$, where $q\in \mathbb{R}$ is the ratio of the geometric progression.  From first Vieta's formulas I get $\alpha q^2+\alpha q+(1+m) = 0$ and from third Vieta's formulas I get $q\alpha = \sqrt[3]{-1}$. From here I stuck. I don't know if my way of working this out is the right way. If it is could you please help me complete the solution, and if not I would very much appreciated If you would provide me a solution for this exercise. 
 A: So you have
\begin{align*}
\alpha(1+q+q^2)&=-m\\
\alpha^2q(1+q+q^2)&=-3\\
(\alpha q)^3&=-1
\end{align*}
So $\alpha q, 1+q+q^2\neq 0$.  Thus dividing the second by the first gives $\alpha q=\frac3m\in\mathbb{R}$, so $\alpha q=-1$.  Then the second equation gives
$$
1+q+q^2=-3q
$$
which you can solve for $q=(2+\sqrt3)^{\pm 1}$ and hence $\alpha$.
A: I would write for the three members of a geometric progression:
$$a_1,a_1q,a_1q^2$$ then you will get
$$a_1^3+ma_1^2-3a_1+1=0$$
$$a_1q^3+ma_1^2q^2-3a_1q+1=0$$
$$a_1^3q^6+ma_1^2q^4-3a_1q+1=0$$
and
$$-(a_1+a_1q+a_1q^2)=m$$
$$a_1^2q+a_1^2q^2+a_1^2q^3=-3$$
$$-a_1^3q^3=1$$
Eliminating $a_1$ we get $$q+q^2+q^3=-3q$$
and $$1+q+q^2=mq$$ for $q\neq 0$
A: Assume the roots of the cubic are: 
$$\frac{a}{r}, a, ar$$
Because it is given the roots need to be in GP. Product of the roots is given to be $-1$
$$\implies a^3 = -1$$
$$\implies a = -1$$
Also is given sum of products taken two at a time to be $-3$. 
$$\implies \frac{1}{r} + r + 1 = -3$$
$$\implies r^2 +4r +1 = 0$$
Solve for $r$. And then you'll be able to solve for $m$.
