<s>[INST] Find $x^5+x^{-5}$ given the value of $x^2+x^{-2}$. 
Find $x^5+\dfrac1{x^5}$ in its simplest form given that $x^2+\dfrac1{x^2}=a$ for $a,x>0$.

Attempt:
We write $$x^2+\frac1{x^2}=a\implies x^4-ax^2+1=0\implies x^5=ax^3-x$$ and $$\frac1{x^2}=a-x^2\implies \frac1{x^4}=a^2-2ax^2+x^4\implies \frac1{x^5}=\frac{a^2}x-2ax+x^3$$ so $$\begin{align}x^5+\frac1{x^5}&=(1+a)x^3-(1+2a)x+a^2\cdot\frac1x\\&=(1+a)x^3-(1+2a)x+a^2(ax-x^3)\\&=(1+a-a^2)x^3-(1+2a-a^3)x\\\implies x^5+\frac1{x^5}&=(a^2-a-1)x(a+1-x^2)\end{align}$$ But is this in its simplest form? 
 [/INST] With $S_n:=x^n+x^{-n}$, you can establish the recurrence
$$S_{n+1}=S_nS_1-S_{n-1}.$$
From this, $$S_2=S_1^2-2,
\\S_3=S_1(S_2-1),
\\S_4=S_2^2-2,
\\S_5=S_1(S_2^2-S_2-1).
$$
You draw $S_1=\sqrt{S_2+2}$ from the first identity.
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