+69 Actualy rationalizing the denominator means getting rid of the roots, you did not do that.
+118687 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
