$x^3-3x-3=0$, prove that $10^x<127$

As suggested by the comments, approximations for $x$ may be found in terms of continued fractions. Then compare with continued fraction approximations for $\ln\frac{40}{237}$ (obtained form the Taylor expansion?)

By checking signs at $x=-1$ and $x_=-2$ we see that there is a real root in $]-2,-1[$. Now substitute $x\leftarrow \frac1y-2$ and multiply by $y^3$ to find $$ -6y^3+31y^2-18y+3$$ as new polynomial. Verify by sign changes that there is a root between $y=4$ and $y=5$. Then substitue $y\leftarrow \frac1z+4$ etc.

Unfortunately, the continued fraction of the root is $[-2, 4, 1, 1, 8, 4, 11, 5,\ldots]$ and that of $\ln\frac{40}{237}$ is $[-2, 4, 1, 1, 8, 4, 11, 8,\ldots]$, so there are quite a few laborious steps in front of you.