Can We Evaluate A Nice Radical

[@mathstuff-n5u]

thanks today i manage to generalize formula for equation : x^3+bx+c = 0 using same idea

[@MichaelRothwell1]

Very elegant solution of the cubic x³-x-1=0 using Cardano's method!
We can easily use this method to find all three roots of the cubic.
Having found a³=(9+√(69))/18, then from a³+b³=1 we get b³=(9-√(69))/18.
For each possible value of a, we get the corresponding value of b by choosing the cube roots so their product is real, as we have ab=1/3.
The result is:
a=³√((9+√(69))/18), b=³√((9-√(69))/18) (real solutions)
a=³√((9+√(69))/18)ω, b=³√((9-√(69))/18)ω²,
a=³√((9+√(69))/18)ω², b=³√((9-√(69))/18)ω
where ω=cis(2π/3)=-1/2+√3/2i, the complex primitive cube root of 1, with the solutions being a+b in each case.
Convergence of the sequence ³√1, ³√(1+³√(1)), ³√(1+³√(1+³√(1))),... is straightforward to prove: it is easy to show by induction both that the sequence is increasing, and that each term is less than 2, so the result follows by the monotone convergence theorem.

[@alexandermorozov2248]

Nice solution! 👍

[@ShortsOfSyber]

Thanks! 😊

[@phill3986]

👍😀✌️👏😎👏✌️😀👍

[@ZIN24031980]

Nice solution, I'd liked it.

[@ShortsOfSyber]

Thanks! 😊

[@GuiveChafai]

you did not Prove that the limit exists. You assumed that the limit exists and then calculated.

[@hitman6540]

What! There isn't a limit in this video bro

[@MichaelRothwell1]

Indeed. Please see my comment for a sketch of the proof of convergence.