<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] As OP noted, $\,x = 2 + \sqrt{3}\,$, and also $\,\dfrac{1}{x} = 2 - \sqrt{3}\,$. Then $\,\color{blue}{x+ \dfrac{1}{x} = 4}\,$, $\,\color{green}{x - \dfrac{1}{x} = 2\sqrt{3}}\,$, and:
$$
\require{cancel}
\begin{align}
& \;\;x^6 - \color{green}{2\sqrt{3}}\, x^5 - x^4 + x^3 - \color{blue}{4}\,x^2 + 2x - \sqrt{3}
\\ = &\; \cancel{x^6} - \left(\color{green}{\cancel{x} - \bcancel{\frac{1}{x}}}\right)x^5 - \bcancel{x^4} + \xcancel{x^3} - \left(\color{blue}{\xcancel{x}+\frac{1}{x}}\right)x^2 + 2x - \sqrt{3}
\\ = &\;\; x - \sqrt{3}
\\ = &\; \;2
\end{align}
$$
</s>