Well, there was this from a related issue, but I thought the specifics got
in the way of what we really want to show. The user, I think, will
recognize when they've got some cruft to chop. But I can include it if you
want.

>>> v=S('''-1484/375 - (-2 + 2*sqrt(3)*I)*(-12459439/52734375 +
...     114*sqrt(12657)/78125)**(1/3) - 172564/((-140625/2 +
...     140625*sqrt(3)*I/2)*(-12459439/52734375 +
...     114*sqrt(12657)/78125)**(1/3))''')
>>> v.n()
0.655814631227503 - 0.e-22*I
>>> v.n(chop=True)
0.655814631227503

This might work, too:

>>> (1+I*(cos(1)**2+sin(1)**2-1)).n()
1.0 - 0.e-71*I
>>> (1+I*(cos(1)**2+sin(1)**2-1)).n(chop=True)
1.00000000000000

/c