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> >
> > can someone help me to verify that the solutions of the cubic
> > ax^3 + (a - 1)x^2 - (b + 1)x + 1 - 2b = 0
> >
> > can be defined as follows:
> >
> > say b=na+m,
> > and y1=x2-x1,y2=x3-2x2+x1 (x1,x2,x3 are the roots of the cubic)
> >
> > then the following quadratic form in y1,2
> > (2n+1)a^2+(3m+1)a+1=3y1^2+3y1y2+y2^2
> > holds.
> >
Wrong in general.
Take a case where everything is rational...
x1 = -3, x2 = -2, x3 = 0
a = -1/4, b = 1/2
n = 2, m = 1
Then
ax^3 + (a - 1)x^2 - (b + 1)x + 1 - 2b = a(x-x1)(x-x2)(x-x3)
y1 = 1, y2 = 1
but
(2n+1)a^2+(3m+1)a+1 = 3y1^2+3y1y2+y2^2
fails.
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