<s>[INST] Value of $x^6 - 2\sqrt{3}x^5 - x^4 + x^3 - 4x^2 + 2x - \sqrt{3}$ given $x = \frac{1}{2-\sqrt{3}}$? It is given that $x = \frac{1}{2-\sqrt{3}}$. Find the value of $x^6 - 2\sqrt{3}x^5 - x^4 + x^3 - 4x^2 + 2x - \sqrt{3}$.
Well I tried rationalising and I came to know that $x = 2 + \sqrt{3}$. And I know that directly putting the values wont help either but I am not able to factorize the polynomial or manipulate it to help me.
I would be grateful if anybody could help me.
 [/INST] I calculated it longhand:
$$ \begin{align} x & = 2 + \sqrt 3 \\ x^2 & = 2^2 + 3 + 2 \cdot 2 \sqrt 3 \\ & = 7 + 4 \sqrt 3 \\ x^3 & = 2 \cdot 7 + 4 \cdot 3 + 2 \cdot 4 \sqrt 3 + 7 \sqrt 3 \\ & = 26 + 15 \sqrt 3 \\ x^4 & = 97 + 56 \sqrt 3 \\ x^5 & = 362 + 209 \sqrt 3 \\ x^6 & = 1351 + 780 \sqrt 3 \end{align} $$
We also have
$$ 2 \sqrt 3 x^5 = 209 \cdot 6 + 362 \cdot 2 \sqrt 3 = 1254 + 724 \sqrt 3 $$
Then
$$ \begin{align} & x^6 - 2 \sqrt 3 x^5 - x^4 + x^3 - 4x^2 + 2x - \sqrt 3 \\ & = 1351 - 1254 - 97 + 26 - 28 + 4 + (780 - 724 - 56 + 15 - 16 + 2 - 1) \sqrt 3 \\ & = 2 \end{align} $$
It was only after doing so that I noticed @JoséCarlosSantos's short cut.
</s>