$x^3-3x-3=0$, prove that $10^x<127$ $x$ is the real root of the equation $$3x^3-5x+8=0,\tag 1$$ prove that $$e^x>\frac{40}{237}.$$
I find this inequality in a very accidental way,I think it's very difficult,because the actual value of $e^x$ is $0.1687763721…$,and the value of $\frac{40}{237}$ is $0.1687763713…$
I have no idea on this problem,I just know that the equation $(1)$ has only one real root.Thank you!
$(Edit)$A similar problem: $x$ is the real root of the equation $$x^3-3x+3=0,\tag 2$$ prove that $$10^x>\frac{1}{127}.$$

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.
