HP50g simplifing a root

10062020, 09:40 AM
Post: #15




RE: HP50g simplifing a root
(10062020 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 = (xy)/3, where x=A/a, y=a*a B = b * (3a² + b²k) R = B²k = r * (3a² + r)^2 = (xy)/3 * (3y + (xy)/3)² = (xy)/27*(x+8*y)² XCas> getR(x,y) := (x+8*y)^2*(xy)/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/aa*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) → \([90+34\sqrt7,\; 90+101i\sqrt7,\; 90+602i\sqrt{73}, \;90+2401i \sqrt{299}]\) 

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