## Rekation Between Coefficients For a Cubic When Roots Form a Geometric Sequence

If the roots of a cubic equation
$ax^3+bx^2+cx+d=0$
form a geometric progression is there a relationship between the coefficients?
Yes.
Let the roots be
$\alpha , \; \beta , \; \gamma$
and divide the equation above by
$a$
to give
$x^3 + \frac{b}{a}x^2+ \frac{c}{a}x+ \frac{d}{a}$
.
We can write
$(x- \alpha )(x- \beta )(x- \gamma )=x^3+x(- \alpha - \beta - \gamma_)+x^2(\alpha \beta +\alpha \gamma + \beta \gamma ) +(-\alpha \beta \gamma )=0$

Equating coefficients gives
$\frac{b}{a}=-(\alpha + \beta +\gamma )$
(1)
$\frac{c}{a}= \alpha \beta + \alpha \gamma + \beta \gamma$
(2)
$\frac{d}{a}=- \alpha \beta \gamma$
(3)
Since the roots form a geometric sequence we can write
$\alpha = x, \; \beta =xy, \; \gamma =xy^2$
.
Using these in (1), (2) and (3) gives
$\frac{b}{a}=-(x+xy+xy^2)=-x(1+y+y^2)$
(4)
$\frac{c}{a}= x^2y +x^2y^2+x^2y^3=x^2y(1+y+y^2)$
(5)
$\frac{d}{a}=- x^3y^3$
(6)
(5) divided by (4) gives
$\frac{c}{b}=xy \rightarrow (\frac{c}{b})^3=-(xy)^3=\frac{d}{a}$
.
Hence
$ac^3=b^3d$
.