A Golden Question

$\begin{aligned} (w-1)(w+1) & = w \\ \Rightarrow w^2 - 1 & = w \\ w^2 -1 -w & = 0 \\ \Rightarrow w - w^{-1} - 1 & = 0 \\ w - w^{-1} & = 1 \\ & \\ \color{#3D99F6}{w^2 + w^{-2}} & = \left(w - w^{-1}\right)^2 + 2 = 1 + 2 = \color{#3D99F6}{3} \\ & \\ \left(w - w^{-1}\right)^3 & = w^3 - 3w + 3w^{-1}-w^{-3} \\ \Rightarrow 1 & = w^3 - 3(1) -w^{-3} \\ \Rightarrow w^3 -w^{-3} & = 4 \\ \left(w^3 -w^{-3}\right)^2 & = w^6 -2 + w^{-6} \\ \Rightarrow \color{#D61F06}{w^6 + w^{-6}} & = 4^2 + 2 = \color{#D61F06}{18} \\ & \\ \left(w^2 + w^{-2}\right)^5 & = w^{10} + 5w^6 + 10w^2 + 10w^{-2} + 5w^{-6} + w^{-10} \\ \Rightarrow w^{10} + w^{-10} & = \left(\color{#3D99F6}{w^2 + w^{-2}} \right)^5 - 5 \left(\color{#D61F06}{w^6 + w^{-6}} \right) - 10 \left( \color{#3D99F6}{w^2 + w^{-2}} \right) \\ & = \color{#3D99F6}{3}^5 - 5(\color{#D61F06}{18}) - 10(\color{#3D99F6}{3}) = \boxed{123} \end{aligned}$

$(w-1)(w+1)=w$ $w^2=w+1$ $w=\varphi,-\varphi^{-1}$ we see that puting both in the given express will give the same answer. here, $\varphi$ is the golden ratio. we plug in either of these values and the expression becomes $\varphi^{10}+(-\varphi)^{-10}$ we know $L_n=\varphi^{n}+(-\varphi)^{-n}$ where $L_n=\begin{cases} 2,\quad n=0\\1,\quad n=1\\ L_{n-1}+L_{n-2} ,\quad n>1\end{cases}$ . we can easily find the 10th lucas numer. $2,1,(1+2=)3,(3+1=)4,(4+3=)7,(7+4=)11,(11+7=)18,(18+11=)29,(29+18=)47,(47+29=)76,(76+47=)\boxed{123}$