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  
\[\frac{c}{a}= \frac{2b^2}{9a^2}+ \frac{3d}{b}\]
.

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