## Rekation Between Coefficients For a Cubic When Roots Form an Arithmetic Sequence

If the roots of a cubic equation
$ax^3+bx^2+cx+d=0$
form an arithmetic 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 an arithmetic sequence we can write
$\alpha = x-y, \; \beta =x, \; \gamma =x+y$
.
Using these in (1), (2) and (3) gives
$\frac{b}{a}=-(x-y+x+x+y)=-(3x)$
(4)
$\frac{c}{a}=(x-y)x+(x-y)(x+y)+x(x+y)= 3x^2-y^2$
(5)
$\frac{d}{a}=-(x-y)x)x+y)=-x(x^2-y^2)$
(6)
We can write (5) as
$\frac{c}{a}= 2x^2+x^2-y^2$
.
From (4)
$x=- \frac{b}{a} \rightarrow 2x^2= \frac{2b^2}{9a^2}$
.
From (4) and (6)
$(x^2-y^2)= - \frac{1}{x} \times - \frac{d}{a} = \frac{3a}{b} \frac{d}{a}=\frac{3d}{b}$
.
Hence
$\frac{c}{a}= \frac{2b^2}{9a^2}+ \frac{3d}{b}$
.