HP50g simplifing a root

[Albert Chan]

For general case, to solve for all a, b:

      ³√(n + m√k) = a + b√k

(a + b√k)³ = a³ + 3a²b√k + 3ab²k + b³k√k

n = a³ + 3ab²k       → n/a - a² = 3b²k
m = 3a²b + b³k      → 3m/b - 9a² = 3b²k

Equate the 2 to eliminate k, we have b = 3ma / (n + 8a³)

Substitute b to n = ... equation, we have a cubic equation, as function of A = a³

XCas> f(n, m, k) := horner([64, -48*n, 27*k*m^2-15*n^2, -n^3], A)

For first example, ³√(26 - 15√3):

XCas> expand(f(26,-15,3))       → 64*A^3 - 1248*A^2 + 8085*A - 17576
XCas> proot(ans())                  → [5.75-1.125*i , 5.75+1.125*i , 8.0]

Above solved for A = a³. Each A produce 3 roots of a, thus a totaled 3×3 = 9 roots.
For this example, the real root is integer, a=2.

XCas> b := 3*m*a/(n + 8*a^3)
XCas> subst([a, b], [n,m,a] = [26,-15,2])       → [2, -1]