Let P(x)=ax^3+bx^2+cx+d from R[X] with a<>0.If the coeficients of

> this polinomial are in arithmecical progression, and all of its

roots

> are real then the following relations holds:

> arctan(1/x1)+arctan(1/x2)+1/(arctan(x3))=-pi/4.

where x1,x2,x3 are its roots.

Suggested solution:

arctan(1/x1)+arctan(1/x2)=arctan[(1/x1+1/x2)/(1-1/x1x2)]=

arctan[(x1+x2)/(x1x2-1)]

Adding arctan(1/x3)

arctan(1/x1)+arctan(1/x2)+arctan(1/x3)=

arctan[(x1x3+x2x3+x2x1-1)/(x1x2x3-x3-x1-x2)](1)

From the theory of equations:

x1+x2+x3=-b/a

x1x2+x2x3+x1x3=c/a

x1x2x3=-d/a

thus replacing we have in (1):

x1x3+x2x3+x2x1-1=c/a-1=(c-a)/a

x1x2x3-x3-x1-x2=-d/a+b/a=(b-d)/a

But since a,b,c, d are in AP

a=A-3D

b=A-D

c=A+D

d=A+3D

c-a=4D

b-d=-4D

Thus (1) reduces to

arctan[(c-a)/(b-d)]=arctan[4D/-4D]=-pi/4

Note that the solution of arctan must lie within -pi/2 to pi/2

Saurabh