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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 $beta/r,beta, rbeta$ is
$$ x^3-betaleft(1+r+frac1rright) x^2+beta^2left(1+r+frac1rright)x-beta^3$$
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?

If $alpha, beta, gamma$ are the roots of $ax^3+bx^2+cx+d=0$

Using Vieta's Formulas, $$alpha+beta+gamma=-frac ba, alphabeta+betagamma+gammaalpha=frac ca,alphabetagamma=-frac da$$

If $alpha,beta,gamma$ are in Geometric Progression, $fracbetaalpha=fracgammabeta$ is constant $=r$(say).
Clearly, $alpha rne 0$

$implies beta=alpha r, gamma=beta r=alpha r^2$

$$implies alpha+alpha r+alpha r^2=-frac baimplies alpha(1+r+r^2)=-frac ba$$

$$alphaalpha r+alpha ralpha r^2+alpha r^2alpha=frac caimplies alpha^2 r(1+r+r^2)=frac ca$$

$$alphaalpha ralpha r^2=-frac daimplies alpha^3 r^3=-frac da$$

So, $$frac{alpha^2 r(1+r+r^2)}{alpha(1+r+r^2)}=frac{frac ca}{-frac ba}implies alpha r=-frac cb text{ if }1+r+r^2ne0$$

Then, $$left(-frac cbright)^3=-frac daimplies b^3d=c^3a$$

If $1+r+r^2=0,$

$b=c=0$
and $r=frac{-1pmsqrt3}2$ i.e., r is a complex cube root of $1$

So, the equation reduces to $ax^3+d=0$ whose roots are $left(-frac daright)^frac13,left(-frac daright)^frac13r, left(-frac daright)^frac13r^2$.

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 $y=alpha+beta=alpha+beta+gamma-gamma=-frac ba-gammaimplies gamma=-left(y+frac baright)$

As $gamma$ is a root of the given equation, $-aleft(y+frac baright)^3+bleft(y+frac baright)^2-cleft(y+frac baright)+d=0$

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, $(alpha+beta)(beta+gamma)(gamma+alpha)=0implies ad=bc$ (the condition is necessary)

Sufficiency: Again, if $ad=bc,$ $frac ab=frac cd=t$(say,) so $a=bt,c=dt$

Then the equation becomes $$btx^3+bx^2+dtx+d=0implies bx^2(tx+1)+d(tx+1)=0$$

So, $$(tx+1)(bx^2+d)=0$$ One root is $-frac1t$ What about the others?

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.
$$q^3-p^3r \ =(alpha beta + beta gamma + alpha gamma)^3-alpha beta gamma(alpha + beta + gamma) \ = alpha^3beta^3+alpha^3gamma^3+beta^3gamma^3-alpha beta gamma(alpha^3+beta^3+gamma^3) \ = (alpha^2-beta gamma)(beta^2-alpha gamma)(gamma^2-alpha beta) \=0$$
So the roots are in geometric progression.