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I am getting tired of using expansion to solve this problem. I wonder if there is any non-algebra ways to solving it.

Problem: Suppose that the number $x$ satisfies the equation $x+x^{-1}=3$. Compute the value of $x^7+x^{-7}$.

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    @sasha: A simple way is to verify the recurrence mentioned above by Jyrki Lahtonen. The first two $$P$_$k$$ are integers, and then by the recurrence they all are.2011-10-20

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Here is another way which uses the Lucas Numbers:

We can rearrange the equality $x^{-1}+x=3$ to get the polynomial $x^2-3x+1=0$. Notice that the only solutions are $x=\phi^2$ and $x=\phi^{-2}$ where $\phi=\frac{1+\sqrt{5}}{2}$ is the golden ratio. (In particular one root is $\frac{3+\sqrt{5}}{2}=\left(\frac{1+\sqrt{5}}{2}\right)^2$)

Since for either choice $x=\phi^2$ or $x=\phi^{-2}$, the quantity $x^n+x^{-n}$ is the same, we can write $x^n+x^{-n}=\phi^{2n}+\phi^{-2n}.$ This above sum has a particular name, and is known as the $2n^{th}$ Lucas Number. It can be described by a recurrence relation like the Fibonacci numbers. In particular we have that $L_n=\phi^n+\phi^{-n}.$ For example this gives us that $x^{8}+x^{-8}=L_{16}=2207.$ Similarly $x^{-7}+x^7=L_{14}=843.$

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Edit: The original question asked to evaluate $x^8+x^{-8}$, but was edited by the OP after this answer was composed.

I am not sure if this counts as a new way, but notice that in general $\left(u^{-1}+u\right)^2=u^{-2}+u^2+2.$

This means that from the first equation we can read off $u^{-2}+u^2 =9^2-2=7$ and then $u^{-4}+u^4 =7^2-2=47$ and lastly $u^{-8}+u^8 =47^2-2=2207.$

Remark: This was actually the key idea in the solution to B4 on the putnam in 1995. Specifically that problem was

B4: Evaluate $\left(2207-\frac{1}{2207-\frac{1}{2207-\cdots}}\right)^{\frac{1}{8}}$ and write your answer in the form $\frac{a+b\sqrt{c}}{d}$ where $a,b,c,d$ are integers.

The solution is given by letting $x$ equal the above quantity. Taking 8th powers and rearranging, we notice that $x^8+x^{-8}=2207.$ From here we do the reverse of my above steps to conclude $x$ is one of the roots of $x+x^{-1}=3$. (Or in other words, the golden ratio squared)

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Ok. Hint for the original question about $x^8+x^{-8}$: $ x^2+2+\frac1{x^2}=(x+\frac1x)^2=3^2=9 \Rightarrow x^2+\frac1{x^2}=7. $ $ x^4+2+\frac1{x^4}=(x^2+\frac1{x^2})^2=7^2=49 \Rightarrow \ldots $