Algebra

${ \ {1} \over 3 root 2 +4}\\$

Actualy rationalizing the denominator means getting rid of the roots, you did not do that.

The sum of 2 cubics is  ..  (something you need to memorize)

$a^3+b^3=(a+b)(a^2-ab+b^2)$

So if you have a denominator of  $(a+b)$ where a or b have cube roots, and you multiply it by  $(a^2-ab+b^2)$ then you will end up with a rational denominator.

$\frac{1}{\sqrt[3]{2}+\sqrt{16}}=\frac{1}{\sqrt[3]{2}+4}$

let

$a= 2^{1/3}\qquad    and\qquad b=4\\so\\ a^2=2^{2/3}\qquad    and\qquad b^2=16\\ ab=4*2^{1/3}$

$a^3+b^3=(a+b)(a^2-ab+b^2)$

$\frac{1}{\sqrt[3]{2}+4}\times \frac{2^{2/3}-4*2^{(1/3)}+4^2}{2^{2/3}-4*2^{(1/3)}+4^2}\\~\\ =\frac{2^{2/3}-4*2^{(1/3)}+4^2}{2+64}\\~\\ =\frac{\sqrt[3]{4}-4\sqrt[3]{2}+16}{66}\\~\\$

Here is a video that might help you understand

https://www.youtube.com/watch?v=33xwVWyrFgg