Find the Frenet Frame of $\alpha(t)=(2t,t^2,\frac{t^3}{3})$ for $t\in\mathbb{R}$
First finding $\alpha'=(2,2t,t^2)$ and $\alpha''=(0,2,2t)$
$||\alpha'||=\sqrt{4+4t^2+t^4}=\sqrt{(t^2+2)^2}=t^2+2 \neq 1$ so not arc length parameterized
$T=\frac{\alpha'}{||\alpha'||}=(\frac{2}{t^2+2}, \frac{2t}{t^2+2}, \frac{t^2}{t^2+2})$
$N=\frac{T'}{||T'||}$
However I am struggling from here:
$T'=(\frac{-4t}{(t^2+2)^2}, \frac{4-2t^2}{(t^2+2)^2}, \frac{4t+2t^3-2t^4}{(t^2+2)^2})$
$$||T'||=\sqrt{\frac{16t^2+16-16t^2+4t^3+16t^2+8t^4-8t^5+8t^4+4t^6-4t^7-8t^5-4t^7+4t^8}{(t^2+2)^4}}=\sqrt{\frac{16+20t^2+16t^4-16t^5+4t^6-8t^7+4t^8}{(t^2+2)^4}}$$
Providing my calculation is correct, how do I simplify this crazy expression?
So using $B=\frac{\alpha' \times \alpha''}{||\alpha' \times \alpha''||}$
$\alpha' \times \alpha'' = (2t^2, -4t, 4)$
$||\alpha' \times \alpha''||=\sqrt{4t^4+16t^2+16}=2\sqrt{(t^2+2)^2}=2(t^2+2)$
$B=(\frac{t^2}{t^2+2}, \frac{-2t}{t^2+2}, \frac{2}{t^2+2})$
$N=-(T \times B)=(\frac{-2t^3-4t}{(t^2+2)^2}, \frac{-t^4+4}{(t^2+2)^2}, \frac{2t^3+4t}{(t^2+2)^2})$