<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] Note that\begin{align}x^6-2\sqrt3x^5&=x^5\left(x-2\sqrt3\right)\\&=x^5\left(2-\sqrt3\right)\\&=x^4x\left(2-\sqrt3\right)\\&=x^4\end{align}and that therefore$$x^6-2\sqrt3x^5-x^4+x^3-4x^2+2x-\sqrt{3}=x^3-4x^2+2x-\sqrt3.$$Now, note that\begin{align}0&=\left(x-\left(2+\sqrt3\right)\right)\left(x-\left(2-\sqrt3\right)\right)\\&=x^2-4x+1.\end{align}So,\begin{align}x^3-4x^2+2x-\sqrt3&=x\overbrace{(x^2-4x+1)}^{\phantom0=0}+x-\sqrt3\\&=x-\sqrt3\\&=2.\end{align}
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