I thought the problem would go away if you shifted away from zero (where the floating point number representation can shift almost arbitrarily), but it still cannot do any of:

Code:

solve(x=-1, 2, x^3)
solve(x=-1+0.1, 2+0.1, (x-0.1)^3) - 0.1
solve(x=-1+0.1, 2+0.1, print(x);(x-0.1)^3) - 0.1

with usual realprecision default. /JeppeSN