so..., someone in here asked about how to rationalize higher index roots, they sad something about the telescoping identity, but I don't get it. I know there is another way to do it and has something to do with the ruffini's rule.
please some help
$$\frac{1}{\sqrt[10]{2}-1}$$
Let us consider in general the polynomial $P(x)=x^{10}-1$ (your problem is the particular case $x=\sqrt[10]{2}$). Now you see that, since $P(1)=0$ this means that you can factor $P(x)=(x-1)\,Q(x)$. You can use Ruffini to find $Q$.
Multiply numerator and denominator by $Q$ so that: $$ \frac{1}{x-1}=\frac{Q(x)}{(x-1)\,Q(x)}=\frac{Q(x)}{x^{10}-1} $$
Substitute the value $x=\sqrt[10]{2}$ and the root is removed from the denominator.