Verification of Frenet Serret I'm trying to show that (1) $T'\times T'' = k^2(kB +\tau T)$       
$T' = \kappa N$, from Frenet Serret
$T'' = \kappa'N + N'\kappa$, but the algebra didn't follow when I tried to substitute this on the Left hand side, of (1) above
 A: First using the frenet serret equation $$N' = -kT + \tau B,$$ substitute it into $T''$ to get
$$T'' = k'N - k^2 T + k\tau B$$
so $$T' \times  T'' = kN \times  (-k'N - k^2T + kτB)
= -kk'(N \times  N) - k^3(N \times T) + k^2τ(N \times B)$$
If you unsure how I got to this point, here's a link to the properties of the cross product
Since $\{T,N,B\}$ is an orthonormal basis (meaning each vector is of unit length and each vector is orthogonal to one another), $$N \times N = 0,\quad  T \times N = B$$ which implies $$N \times T = -B,\quad N \times B = T.$$
Going back to our equation we get
$$T' \times T'' = k^3B + k^2\tau T = k^2(kB + \tau T)$$
A: Not really much to add after Kevin's rightfully accepted answer (+ 1), but the derivation is so cute I wanted to write it up for myself.
The first Frenet-Serret equation,
$\dot T = \kappa N, \tag 1$
may be differentiated with respect to the arc-length $s$ to give
$\ddot T = \dfrac{d}{ds}(\kappa N) = \dot \kappa N + \kappa \dot N \tag 2$
the second Frenet-Serret equation,
$\dot N = -\kappa T + \tau B, \tag 3$
may be substituted into (2):
$\ddot T = \dot \kappa N + \kappa (-\kappa T + \tau B) = \dot \kappa N -\kappa^2 T + \kappa \tau B; \tag 4$
we take the cross product of (1) and (4):
$\dot T \times \ddot T = \kappa N \times (\dot \kappa N -\kappa^2 T + \kappa \tau B) = \kappa \dot \kappa N \times N - \kappa^3 N \times T +  \kappa^2 \tau N \times B$
$= - \kappa^3 N \times T +  \kappa^2 \tau N \times B; \tag 5$
$B = T \times N \Longrightarrow N \times T = -B, \; N \times B = T; \tag 6$
thus (5) yields
$\dot T \times \ddot T = \kappa^3 B + \kappa^2 \tau T = \kappa^2 (\kappa B + \tau T), \tag 7$
as per request.  QED.
