Amir Hossein wrote: Let $t$ be a positive real number such that $t^4 + t^{-4} = 2023$. Determine the value of $t^3 + t^{-3}$ in the form of $a\sqrt b$, where $a$ and $b$ are positive integers. Let $s_k=t^k+t^{-k}$ We have $s_4=2023$ So $s_2^2=s_4+2=2025$ and $s_2=45$ So $s_1^2=s_2+2=47$ and $s_1=\sqrt{47}$ And $s_1s_2=s_3+s_1$ gives $\boxed{s_3=44\sqrt{47}}$

$t^4+t^{-4}=2023$ $t^4+2t^2t^{-2}+t^{-4}=2023+2$ $(t^2+t^{-2})^2=2025$ $t^2+t^{-2}=45$ $t^2+2t^1t^{-1}+t^{-2}=45+2$ $(t+t^{-1})^2=47$ $t+t^{-1}=\sqrt{47}$ $(t^2+t^{-2})(t+t^{-1})=45\sqrt{47}$ $t^3+t^{-3}+t+t^{-1}=45\sqrt{47}$ $t^3+t^{-3}+\sqrt{47}=45\sqrt{47}$ $t^3+t^{-3}=45\sqrt{47}-\sqrt{47}$ $t^3+t^{-3}=44\sqrt{47}$