Show that the torsion satisfies $\tau(s)=\frac{-\alpha'(s)\wedge\alpha''(s)\cdot\alpha'''(s)}{|\kappa(s)|^2}$ Let $\alpha:I\rightarrow\mathbb{R}^{3}$ a curve parametrized by arc lenght $s$. Denote by $t(s)=\alpha'(s)$ the tangent vector, $\kappa(s)=|\alpha'(s)|$ the curvature, $n(s)=\alpha''(s)/\kappa(s)$ the normal vector, $b(s)=t(s)\wedge n(s)$ the binormal vector. The torsion is the number $\tau(s)$ such that $b'(s)=\tau(s)n(s).$
I need to show that 
$$\tau(s)=\frac{-\alpha'(s)\wedge\alpha''(s)\cdot\alpha'''(s)}{|\kappa(s)|^2}$$.
I already showed that 
$$\frac{-\alpha'(s)\wedge\alpha''(s)\cdot\alpha'''(s)}{|\kappa(s)|^2}n(s)=
\frac{-t(s)\wedge n(s)\cdot\alpha'''(s)}{\kappa(s)}n(s),$$
and, since $b'(s)=t'(s)\wedge n(s)+t(s)\wedge n'(s),$ I need to prove that 
$$\frac{-t(s)\wedge n(s)\cdot\alpha'''(s)}{\kappa(s)}n(s)=t'(s)\wedge n(s)+t(s)\wedge n'(s). $$
How can I do that?
 A: By definition,
$$ \alpha'(s) = t(s) , $$
and differentiating twice,
$$ \alpha''(s) = t'(s) = \kappa(s) n(s) \\
\alpha'''(s) = \kappa'(s) n(s) + \kappa(s) n'(s) . $$
We can find what $n'$ is: 
$$\begin{align}
n \cdot n &= 1 \implies 0 = (n \cdot n)' = n \cdot n' \\
n \cdot t &= 0 \implies 0 = (n \cdot t)' = n' \cdot t + n \cdot t' = n' \cdot t + \kappa n \cdot n = n' \cdot t + \kappa \\
n \cdot b &= 0 \implies 0 = (n \cdot b)' = n' \cdot b + n \cdot b' = n' \cdot b + \tau
\end{align}$$
so $ n' = -\kappa t - \tau b $, one of the Frenet–Serret formulae (note that Wikipedia uses a different, more common convention that $b' = -\tau n$). Thus,
$$ \alpha'''(s) = \kappa'(s) n(s) - \kappa(s)^2 t(s) - \kappa(s)\tau(s)b(s) $$
If we now compute the triple product of the first three derivatives, we find
$$ \alpha'(s) \times \alpha''(s) = \kappa(s) b(s), $$
so
$$ (\alpha'(s) \times \alpha''(s)) \cdot \alpha'''(s) = -\kappa(s)^2 \tau(s), $$
whence the result. The key here is to work either completely in terms of $\alpha$, or completely in terms of $t,n,b$: mixing them tends to be confusing.
A: You can just calculate that $$\alpha'''=\left(\kappa N\right)'=\kappa N'+\kappa' N=\kappa \tau B+\cdots,$$ using the Frenet equations.  We only care about the $B$ component due to the dot product we want to take. Hence, $$(\alpha'\times\alpha'')\cdot\alpha'''=\kappa^2\tau\implies \tau=\frac{(\alpha'\times\alpha'')\cdot\alpha'''}{\kappa^2}.$$ 
EDIT: Typically, one defines $\tau $ by $B'=-\tau N.$ I see now that you are using $B'=\tau N.$ In this case, the Frenet equations have some sign changes. Taking this into account, you get that $\alpha'''=-\kappa\tau B+\cdots,$ and taking the dot product gives $$(\alpha'\times\alpha'')\cdot\alpha'''=-\kappa^2\tau,$$ or $$\tau=-\frac{(\alpha'\times\alpha'')\cdot\alpha'''}{\kappa^2}.$$
