Calculate the cube root of a number

Haskell - 35

c n=(iterate(\x->(x+n/x/x)/2)n)!!99

Example runs:

c 27  =>  3.0
c 64  =>  4.0
c 1  =>  1.0
c 18.609625  =>  2.6500000000000004  # only first 4 digits are important, right?
c 3652264  =>  154.0
c 0.001  =>  0.1
c 7  =>  1.9129311827723892
c (-27)  =>  -3.0
c (-64)  =>  -4.0

Moreover, if you import Data.Complex, it even works on complex numbers, it returns one of the roots of the number (there are 3):

c (18:+26)  =>  3.0 :+ 1.0

The :+ operator should be read as 'plus i times'

SageMath, (69) 62 bytes

However, don't ever believe it will give you the result, it's very difficult to go randomly through all the numbers:

def r(x):
 y=0
 while y*y*y-x:y=RR.random_element()
 return "%.4f"%y

if you didn't insist on truncating:

def r(x):
 y=0
 while y*y*y-x:y=RR.random_element()
 return y

SageMath, 12 bytes, if exp is allowed

Works for all stuff: positive, negative, zero, complex, ...

exp(ln(x)/3)

J: 16 characters

Loose translation of the Haskell answer:

-:@((%*~)+])^:_~

Test cases:

   -:@((%*~)+])^:_~27
3
   -:@((%*~)+])^:_~64
4
   -:@((%*~)+])^:_~1
1
   -:@((%*~)+])^:_~18.609625
2.65
   -:@((%*~)+])^:_~3652264
154
   -:@((%*~)+])^:_~0.001
0.1
   -:@((%*~)+])^:_~7
1.91293

It works like this:

     (-:@((% *~) + ])^:_)~ 27
↔ 27 (-:@((% *~) + ])^:_) 27
↔ 27 (-:@((% *~) + ])^:_) 27 (-:@((% *~) + ])) 27
↔ 27 (-:@((% *~) + ])^:_) -: ((27 % 27 * 27) + 27)
↔ 27 (-:@((% *~) + ])^:_) 13.5185
↔ 27 (-:@((% *~) + ])^:_) 27 (-:@((% *~) + ])) 13.5185
↔ 27 (-:@((% *~) + ])^:_) -: ((27 % 13.5185 * 13.5185) + 13.5185)
↔ 27 (-:@((% *~) + ])^:_) 6.83313
...

In words:

half =. -:
of =. @
divideBy =. %
times =. *
add =. +
right =. ]
iterate =. ^:
infinite =. _
fixpoint =. iterate infinite
by_self =. ~

-:@((%*~)+])^:_~ ↔ half of ((divideBy times by_self) add right) fixpoint by_self

Not one of the best wordy translations, since there's a dyadic fork and a ~ right at the end.