3th root calculation proof of concept

sqrt3.py

def sqrt3(n, p):
    # any number can be writen as n = a*10 +b, where a >=0, b is 0~9
    # for example 123 = 12 * 10 + 3
    # so then any number N = (a*10 + b)^3 = 1000*a^3 + 300*a^2*b + 30*a*b^2 + b^3 + remainer
    #                                     = 1000*a^3 + b*(300*a^2 + 30*a*b + b^2) + remainer
    # then b*(300*a^2 + 30*a*b + b^2) <= N - 1000*a^3
    # then we can calulate max b to match the expression
    print("sqrt3 {} with {} precision is:".format(n, p))
    if len(n) %3 != 0:
        n = "0"*(3-len(n)%3) +  n 
    splited_n = [n[i:i+3] for i in range(0,len(n),3)]
    A,B,Remainer,Result = 0,0,0,""
    def findb(N, a, b, remainer, result):
        for j in range(0, 11):
            c = remainer*1000 + int(N)
            if j*(300*a*a + 30*a*j +j*j) > c:
                b = j-1
                remainer = c - b*(300*a*a + 30*a*b +b*b)
                a = 10*a + b
                result = "{}{}".format(result, b)
                break
        return a,b,remainer,result
    for i in splited_n:
        A,B,Remainer,Result = findb(i, A, B, Remainer, Result)
    if Remainer != 0:
        Result = "{}.".format(Result)
        for i in ["000" for i in range(p)]:
            A,B,Remainer,Result = findb(i, A, B, Remainer, Result)
    return Result,A,Remainer

> sqrt3("1117", 100)
sqrt3 1117 with 100 precision is:
('10.3757076016152723443705022336612204438590687561861136537248724453317971388480542367950056479536171279',
 103757076016152723443705022336612204438590687561861136537248724453317971388480542367950056479536171279,
 10191420830713470219904633075149898570003885616411667881081123698032858862717609432433416151366991579589177045691626477193310002052568741404591177505785830347460033179729326465418175863355326445000849361)