Edited:what is a way to solve this problem other than use algebra way? 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 $x^7+x^{-7}$.

 A: 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)
A: Ok. Hint for the original question about $x^8+x^{-8}$:
$$
x^2+2+\frac1{x^2}=(x+\frac1x)^2=3^2=9 \Rightarrow x^2+\frac1{x^2}=7.
$$
$$
x^4+2+\frac1{x^4}=(x^2+\frac1{x^2})^2=7^2=49 \Rightarrow \ldots
$$
A: 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 $$x^{-7}+x^7=L_{14}=843.$$  
