If $x+\frac1x=5$ find $x^5+\frac1{x^5}$.
If $x>0$ and $\,x+\dfrac{1}{x}=5,\,$ find $\,x^5+\dfrac{1}{x^5}$.
Is there any other way find it? $$ \left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=23\cdot 110. $$
Thanks
Observe the recurrence relation $$x^{n+1} + x^{-(n+1)} = (x+x^{-1})(x^n+x^{-n}) - (x^{n-1} + x^{-(n-1)}).$$
This immediately gives us the specific recurrence $$f_{n+1} = 5f_n - f_{n-1}, \quad f_0 = 2, \quad f_1 = 5,$$ where $f_n = x^n + x^{-n}$.
\begin{align} x+\frac{1}{x}&=5\\ x^2+1&=5x\\ x^2-5x+1&=0\\ x &=\frac{1}{2} \left( 5 +\sqrt{21}\right)\\ x^5+\frac{1}{x^5}&=\cdots \end{align}