AngouriMath
SolveAPowerOfXOverASumOfTwoCubes(AngouriMath.Entity,AngouriMath.Entity.Variable,System.Boolean)
Method (no overloads)
Summary
A power of x over a sextic that is a sum of two cubes,
d6 ((x^2 + s)^3 + q x^3) , split over the cube roots of unity:
1/(U^3 + V^3) = sum_j w_j/(3 V^2 (U + w_j V)) for U = x^2 + s ,
V = m x , m^3 = q , so that x^k/(U^3 + V^3) is
sum_j w_j x^(k - 2)/(3 m^2 (x^2 + w_j m x + s)) , three quadratics.
Remarks
The sextic is read as d6 x^6 + d4 x^4 + d3 x^3 + d2 x^2 + d0 with
d4^2 = 3 d6 d2 and d2^2 = 3 d4 d0 , which is what makes
d6 x^6 + d4 x^4 + d2 x^2 + d0 the cube d6 (x^2 + s)^3 with
s = d4/(3 d6) . The identity is the partial fractions of 1/(U^3 + V^3) in
U , whose poles -w_j V have residues 1/(3 w_j^2 V^2) = w_j/(3 V^2) ; any
cube rootm of q gives it. The three quadratics are one integral
x^n/(x^2 + B x + S) asked with B and S named. Rubi's 1.3.1
x^k/(27 a^3 + 27 a^2 b x^2 + 27 a^2 c x^3 + 9 a b^2 x^4 + b^3 x^6) was declined,
the integrator's resultant needing numbers for coefficients.
https://github.com/asc-community/AngouriMath/issues/718
cube root
the integrator's resultant needing numbers for coefficients.
https://github.com/asc-community/AngouriMath/issues/718
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