Prove that the equation $x^3-x-1=0$ has only one real root I want to prove that the equation $f(x)=x^3-x-1=0$ has only one real root, which is on the intervall $[1,2]$. 
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I have done the following: 
$f(1)=-1<0$ and $f(2)=5>0$ so $f(1)\cdot f(2)<0$ and so from Bolzano's Theorem we have that the function has at least one root on $[1,2]$. 
We suppose that there are two roots, $a$ and $b$. Then we have that $f(a)=f(b)=0$. 
The function $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$. So, from Rolle's Theorem there is a $c\in (a,b)$ such that $f'(c)=0 \Rightarrow 3c^2-1=0$. 
How can we get a contradiction? 

One can notice, that $f(1)f(2) = -5 < 0$, so there must be root of the equation. Also $f'(x) = 3x^2 - 1$, so it is monotonically increasing in range [1, 2] - there is only one root.
Next step is to prove that $x^3 < x + 1 $ for $x<1$. For $0 < x < 1$ we have $x + 1 > x > x^3$. For $-1 < x < 0$ following is met: $x + 1 > 0 > x ^ 3$. Finally for $x < -1$ we have $1 + x > x > x^3$.
Last step is to prove, that $x^3 > x + 1$ for x > 2, but this is obvious, since $f(2) = 5 > 0$ and for every $x > 2$ $\frac{dx^3}{dx} = 3x^2 > \frac{d(x + 1)}{dx} = 1$.
