HP50g simplifing a root

[Albert Chan]

(09-30-2020 02:22 AM)Albert Chan Wrote:  (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)

Instead of solving cubic, we can simply filter all (a,b), keeping only integers.
Since a divides n, just check divisors of n:

XCas> find_ab(n,m) := remove(x -> frac(x[1]), map(divisors(n).*sign(n) , a->[a, 3*m*a/(n+8*a^3)]))
XCas> find_abk(n,m) := map(find_ab(n,m), x -> x[0] + x[1] * sqrt((n/x[0]-x[0]^2)/(3*x[1]^2)))

XCas> find_abk(26,-15)               → \([2 - \sqrt{3}]\)
XCas> find_abk(9416, -4256)       → \([11 - 7 \sqrt{5}]\)

XCas> simplify((99+100*sqrt(101))^3)        → \(300940299 + 103940300\sqrt{101}\)
XCas> find_abk(300940299, 103940300)      → \([99 + 100 \sqrt{101}]\)