Find the relation between $a, b, c, d$ if the roots of $ax^3+bx^2+cx+d=0$ are in geometric progression.
By considering $(\alpha+\beta)(\beta+\gamma)(\alpha+\gamma)$ show that the above cubic equation has two roots equal in size but opposite in sign if and only if $ad=bc$.
I can do the second part if I am given some hints on the first. I can use $\beta$ as the middle root and make $\alpha=\beta/r$ and $\gamma=r \times \beta$, but haven't got anywhere so far.
The monic polynomial with roots $\beta/r,\beta, r\beta$ is $$ x^3-\beta\left(1+r+\frac1r\right) x^2+\beta^2\left(1+r+\frac1r\right)x-\beta^3$$ We find $\beta=-\frac cb$ and $\beta^3=-\frac da$, hence $$db^3=ac^3$$ as necessary condition. Why is it also sufficient? What additional condition(s) would we need if we wanted the roots to be real?
Suppose $a \not = 0$ , let $p=\frac{b}{a},q=\frac{c}{a},r=\frac{d}{a}$, then the equation is $$x^3+px^2+qx+r=0$$ Using Vieta's Formulas,$$\alpha + \beta + \gamma=-p \\ \alpha \beta + \beta \gamma + \alpha \gamma =q \\ \alpha \beta \gamma = -r$$Suppose $\beta^2=\alpha \gamma$, we get $$q^3=p^3r$$that is $$\left \{ \begin{array}{l} ac^3=b^3d \\ a \not = 0 \\ d \not = 0 \end{array} \right.$$ Now we prove it is sufficient. $$q^3-p^3r \\ =(\alpha \beta + \beta \gamma + \alpha \gamma)^3-\alpha \beta \gamma(\alpha + \beta + \gamma) \\ = \alpha^3\beta^3+\alpha^3\gamma^3+\beta^3\gamma^3-\alpha \beta \gamma(\alpha^3+\beta^3+\gamma^3) \\ = (\alpha^2-\beta \gamma)(\beta^2-\alpha \gamma)(\gamma^2-\alpha \beta) \\=0$$ So the roots are in geometric progression.
If $\alpha, \beta, \gamma$ are the roots of $ax^3+bx^2+cx+d=0$
Using Vieta's Formulas, $$\alpha+\beta+\gamma=-\frac ba, \alpha\beta+\beta\gamma+\gamma\alpha=\frac ca,\alpha\beta\gamma=-\frac da$$
If $\alpha,\beta,\gamma$ are in Geometric Progression, $\frac\beta\alpha=\frac\gamma\beta$ is constant $=r$(say). Clearly, $\alpha r\ne 0$
$\implies \beta=\alpha r, \gamma=\beta r=\alpha r^2$
$$\implies \alpha+\alpha r+\alpha r^2=-\frac ba\implies \alpha(1+r+r^2)=-\frac ba$$
$$\alpha\alpha r+\alpha r\alpha r^2+\alpha r^2\alpha=\frac ca\implies \alpha^2 r(1+r+r^2)=\frac ca$$
$$\alpha\alpha r\alpha r^2=-\frac da\implies \alpha^3 r^3=-\frac da$$
So, $$\frac{\alpha^2 r(1+r+r^2)}{\alpha(1+r+r^2)}=\frac{\frac ca}{-\frac ba}\implies \alpha r=-\frac cb \text{ if }1+r+r^2\ne0$$
Then, $$\left(-\frac cb\right)^3=-\frac da\implies b^3d=c^3a$$
If $1+r+r^2=0,$
$b=c=0$ and $r=\frac{-1\pm\sqrt3}2$ i.e., r is a complex cube root of $1$
So, the equation reduces to $ax^3+d=0$ whose roots are $\left(-\frac da\right)^\frac13,\left(-\frac da\right)^\frac13r, \left(-\frac da\right)^\frac13r^2$.
So, we don't find any relationship among $a,b,c$ and $d$ here.
In the 2nd case, let us find the equation whose roots are $\alpha+\beta,\beta+\gamma, \gamma+\alpha$ using the Transformation of Equations.
If $y=\alpha+\beta=\alpha+\beta+\gamma-\gamma=-\frac ba-\gamma\implies \gamma=-\left(y+\frac ba\right)$
As $\gamma$ is a root of the given equation, $-a\left(y+\frac ba\right)^3+b\left(y+\frac ba\right)^2-c\left(y+\frac ba\right)+d=0$
On simplification, $$ay^3+(\cdots)y^2+(\cdots)y^2+\frac{bc-ad}a=0$$ whose roots are $\alpha+\beta,\beta+\gamma, \gamma+\alpha$
Using Vieta's Formulas again, $(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)=\frac{ad-bc}{a^2}$
If among $\alpha,\beta,\gamma$ two have same absolute value but opposite sign, $(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)=0\implies ad=bc$ (the condition is necessary)
Sufficiency: Again, if $ad=bc,$ $\frac ab=\frac cd=t$(say,) so $a=bt,c=dt$
Then the equation becomes $$btx^3+bx^2+dtx+d=0\implies bx^2(tx+1)+d(tx+1)=0$$
So, $$(tx+1)(bx^2+d)=0$$ One root is $-\frac1t$ What about the others?
