HP50g simplifing a root

[Albert Chan]

(10-06-2020 05:25 AM)peacecalc Wrote:  For calculating A and B I use k as a given value, hmm is that wrong? Are there possibly other solutions with another k as given? But when this would be possibly then we have a lot of more solutinon, too much...?

It is possible for the same A,B, but different k's, we can simplify the cube roots.
(Actually, I am not so sure. Seems pretty rare for this to happen ...)

I know that for same A,k, but different B's, it is possible (at least, if we included imaginary numbers)

Below searched for all A, such that ³√(A ± √R) = a ± √r,       where R = B²k, r = b²k

r = b²k = (A/a - a*a)/3 = (x-y)/3, where x=A/a, y=a*a
B = b * (3a² + b²k)
R = B²k = r * (3a² + r)^2 = (x-y)/3 * (3y + (x-y)/3)² = (x-y)/27*(x+8*y)²

XCas> getR(x,y) := (x+8*y)^2*(x-y)/27
XCas> find_all_a(A) := remove(a-> denom(getR(A/a,a*a))!=1, divisors(A).*sign(A))
XCas> find_all_abk(A) := map(a -> a + sqrt((A/a-a*a)/3), find_all_a(A))

For A=90, k=7, we have 2 cube roots able to simplify, B = 34, 101i
Note: because of "±", we also have the "-" solutions.

XCas> s := find_all_abk(90)  → \([\;3+\sqrt7,\quad\quad 6+i\sqrt7,\quad\;\;
15+i\sqrt{73}, \quad\quad 30+i\sqrt{299}]\)
XCas> simplify(s .^ 3)          → \([90+34\sqrt7,\; 90+101i\sqrt7,\;
90+602i\sqrt{73}, \;90+2401i \sqrt{299}]\)