<s>[INST] Findind the value of $x^4 + 1/x^4$ when $x = 2+3\sqrt{3}$ This question recently came up in a competitive exam and I have been struggling to find out the easiest way to solve it.
Given that $x = 2+3\sqrt{3}$, what is the value of $x^4 + 1/x^4$.
Could someone provide me with an approach other than the direct approach of multiplying $x$ to itself $4$ times and adding it to its reciprocal.
Thanks in advance!
 [/INST] You the thing you need to calculate is $x^4$.
You start by isolating the square root $x = 2+3\sqrt{3} \Leftrightarrow x - 2 = 3\sqrt{3}$
Then you square both sides $(x - 2)^2 = (3\sqrt{3})^2 \Leftrightarrow x^2 - 4x + 4 = 27 \Leftrightarrow x^2 - 4x -23 = 0$
Let $P$ be $X^4$
Let $D$ be $X^2 - 4X - 23$
Now, you compute $R$, the remainder of the euclidean division of $P$ by $D$. I get $R = 248X+897$
If you name $Q$ the quotient, you have $P = D\times Q + R$. You don't need it but I get $X^2+4X+39$
But for your given $x$, $D$ evaluates to $0$ so $P$ evaluates to the same value as $R$. $( X^2+4X+39 ) \times (X^2 - 4X - 23)+ (248X+897) = X^4$
So $X^4 = 1393+744 \sqrt 3$
Then $X^4 + \frac{1}{X^4} = \frac{(X^4)^2+1}{X^4}$ and you finally get $\frac{2 \times (1800529+1036392 \sqrt 3))}{(1393+744\sqrt3)}$
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