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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?

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    It's easier here to use the [alternative strategy](https://en.wikipedia.org/wiki/Frenet%E2%80%93Serret_formulas#Other_expressions_of_the_frame) of calculating $$B = \frac{\alpha'(t) \times \alpha''(t)}{\|\alpha'(t) \times \alpha''(t)\|}.$$Then use that to find $N$. (For what it's worth, it appears your algebra is not quite OK, but it doesn't look very far off, either. Trying to divide the numerator by $t^{2} + 2$ to look for cancellation would be a good strategy if it comes to that.) – 2017-01-10
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    @AndrewD.Hwang How would I then find N from this? – 2017-01-10
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    Since $T \times N = B$, you have $T \times B = -N$. – 2017-01-10
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    @AndrewD.Hwang I have included an answer could you please verify, thanks – 2017-01-11
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    @JohnHughes I have attempted an answer could you check it please – 2017-01-11

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So using $B=\frac{\alpha' \times \alpha''}{||\alpha' \times \alpha''||}$

$\alpha' \times \alpha'' = (2t^2, -4t, 4)$

$||\alpha' \times \alpha''||=\sqrt{4t^4+16t^2+16}=2\sqrt{(t^2+2)^2}=2(t^2+2)$

$B=(\frac{t^2}{t^2+2}, \frac{-2t}{t^2+2}, \frac{2}{t^2+2})$

$N=-(T \times B)=(\frac{-2t^3-4t}{(t^2+2)^2}, \frac{-t^4+4}{(t^2+2)^2}, \frac{2t^3+4t}{(t^2+2)^2})$

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    At a quick check this looks good. Each numerator in $N$ is divisible by $2 + t^{2}$, so you can even simplify a bit further. :) – 2017-01-11
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    A closer look shows that some signs are messed up (or might be). I checked the cross product, and it looks as if only the first term has all the signs correct. Then again, I just woke up 5 min ago, so I could well be wrong. But when I look at the inner product of $T$ with your $N$ (ignoring denominators), I get a $-2t^5$ from the middle term, and another $-2t^5$ from the last term ... and there're no other $t^5$ terms to cancel those. So there must be a sign error. – 2017-01-11
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    @JohnHughes I believe there is an error in the last one, in which is should all be positive. I have changed it now so hopefully it works! – 2017-01-11
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    Yes, although you should probably also simplify following Andrew D. Hwang's comment. – 2017-01-11