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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

$$\frac{1}{\sqrt[10]{2}-1}$$

Blue
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andre neme benedetti
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  • Maybe this helps? https://math.stackexchange.com/questions/3581481/rationalize-this-expression-in-the-description – Hans Lundmark Mar 16 '22 at 22:00
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    with $x = 2^{1/10},$ use $$x^{10} - 1 = (x-1)\left( x^9 + x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 \right) $$ – Will Jagy Mar 16 '22 at 22:11
  • On the other hand, for things other than this special case or similar ones (for example, $\frac{1}{3 \cdot 2^{3/10} + 2^{1/10} - 1}$), as far as I know you need to do something such as linear algebra, or use the extended Euclidean algorithm on polynomials to find a Bezout relation $f(x) (x^{10} - 2) + g(x) (3x^3 + 2x - 1) = 1$. – Daniel Schepler Mar 16 '22 at 22:32
  • @will I would factor further, which us possible with $10$ being composite, thus $x^{10}-1=(x+1)(x-1)C_5(x)C_5(-x)$ where $C_5(u)$ is the fifth cyclotomic polynomial $u^4+u^3+u^2+u+1$. – Oscar Lanzi Mar 16 '22 at 22:48

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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.

Miguel
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