How to compute the nth root of a very big integer

If you are looking for something standard, fast to write with high precision. I would use decimal and adjust the precision (getcontext().prec) to at least the length of x.

Code (Python 3.0)

from decimal import *

x =   '11968003966030964356885611480383408833172346450467339251\
196093144141045683463085291115677488411620264826942334897996389\
485046262847265769280883237649461122479734279424416861834396522\
819159219215308460065265520143082728303864638821979329804885526\
557893649662037092457130509980883789368448042961108430809620626\
059287437887495827369474189818588006905358793385574832590121472\
680866521970802708379837148646191567765584039175249171110593159\
305029014037881475265618958103073425958633163441030267478942720\
703134493880117805010891574606323700178176718412858948243785754\
898788359757528163558061136758276299059029113119763557411729353\
915848889261125855717014320045292143759177464380434854573300054\
940683350937992500211758727939459249163046465047204851616590276\
724564411037216844005877918224201569391107769029955591465502737\
961776799311859881060956465198859727495735498887960494256488224\
613682478900505821893815926193600121890632'

minprec = 27
if len(x) > minprec: getcontext().prec = len(x)
else:                getcontext().prec = minprec

x = Decimal(x)
power = Decimal(1)/Decimal(3)

answer = x**power
ranswer = answer.quantize(Decimal('1.'), rounding=ROUND_UP)

diff = x - ranswer**Decimal(3)
if diff == Decimal(0):
    print("x is the cubic number of", ranswer)
else:
    print("x has a cubic root of ", answer)

Answer

x is the cubic number of 22873918786185635329056863961725521583023133411 451452349318109627653540670761962215971994403670045614485973722724603798 107719978813658857014190047742680490088532895666963698551709978502745901 704433723567548799463129652706705873694274209728785041817619032774248488 2965377218610139128882473918261696612098418

Oh, for numbers that big, you would use the decimal module.

ns: your number as a string

ns = "11968003966030964356885611480383408833172346450467339251196093144141045683463085291115677488411620264826942334897996389485046262847265769280883237649461122479734279424416861834396522819159219215308460065265520143082728303864638821979329804885526557893649662037092457130509980883789368448042961108430809620626059287437887495827369474189818588006905358793385574832590121472680866521970802708379837148646191567765584039175249171110593159305029014037881475265618958103073425958633163441030267478942720703134493880117805010891574606323700178176718412858948243785754898788359757528163558061136758276299059029113119763557411729353915848889261125855717014320045292143759177464380434854573300054940683350937992500211758727939459249163046465047204851616590276724564411037216844005877918224201569391107769029955591465502737961776799311859881060956465198859727495735498887960494256488224613682478900505821893815926193600121890632"
from decimal import Decimal
d = Decimal(ns)
one_third = Decimal("0.3333333333333333")
print d ** one_third

and the answer is: 2.287391878618402702753613056E+305

TZ pointed out that this isn't accurate... and he's right. Here's my test.

from decimal import Decimal

def nth_root(num_decimal, n_integer):
    exponent = Decimal("1.0") / Decimal(n_integer)
    return num_decimal ** exponent

def test():
    ns = "1196800396603096435688561148038340883317234645046733925119609314414104568346308529111567748841162026482694233489799638948504626284726576928088323764946112247973427942441686183439652281915921921530846006526552014308272830386463882197932980488552655789364966203709245713050998088378936844804296110843080962062605928743788749582736947418981858800690535879338557483259012147268086652197080270837983714864619156776558403917524917111059315930502901403788147526561895810307342595863316344103026747894272