Originally Posted by
Soroban Hello, gfbrd!
The original expression has a "minus": .$\displaystyle \sqrt[8]{\frac{2207 \:{\color{red}-}\: 987\sqrt{5}}{2}}$
You must mean: .$\displaystyle \frac{3\:{\color{red}-}\:\sqrt{5}}{2}$
. . which happens to equal: .$\displaystyle \frac{1}{\phi^2}$
. . where $\displaystyle \phi \,=\,\tfrac{1+\sqrt{5}}{2}$, the Golden Ratio.
It can be shown that: .$\displaystyle \frac{1}{\phi^{16}} \:=\:\frac{2207 - 987\sqrt{5}}{2}$
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I'd like to share a "trick" for raising $\displaystyle \phi$ to higher powers.
One of the root of the quadratic equation: .$\displaystyle x^2 - x - 1 \:=\:0$ .is $\displaystyle \phi.$
This means: .$\displaystyle \phi^2 - \phi - 1 \:=\:0 \quad\Rightarrow\quad \boxed{\phi^2 \:=\:\phi + 1}$
Divide by $\displaystyle \phi\!:\;\phi \:=\:1 + \frac{1}{\phi} \quad\Rightarrow\quad \boxed{\frac{1}{\phi} \:=\:\phi - 1}$
We have: .$\displaystyle \frac{1}{\phi} \;=\;\phi - 1$
Square: .$\displaystyle \frac{1}{\phi^2} \;=\;(\phi - 1)^2 \;=\;\phi^2 - 2\phi + 1 \;=\;(\phi+1) - 2\phi + 1 \;=\;2 - \phi$
Square: .$\displaystyle \frac{1}{\phi^4} \;=\;(2-\phi)^2 \;=\;4 - 4\phi + \phi^2 \;=\;4 - 4\phi + (\phi + 1) \;=\;5-3\phi$
Square: .$\displaystyle \frac{1}{\phi^8} \;=\;(5-3\phi)^2 \;=\;25 - 30\phi + 9\phi^2 \;=\;25 - 30\phi + 9(\phi+1) \;=\;34 - 21\phi$
Square: .$\displaystyle \frac{1}{\phi^{16}} \;=\;(34-21\phi)^2 \;=\;1156 - 1428\phi + 441\phi^2 \;=\; 1597 - 987\phi$
Hence: .$\displaystyle \frac{1}{\phi^{16}} \;=\;1597 - 987\left(\frac{1+\sqrt{5}}{2}\right) \;=\;\frac{2207 - 987\sqrt{5}}{2}$
Therefore: .$\displaystyle \sqrt[8]{\frac{2207 - 987\sqrt{5}}{2}} \;=\;\frac{1}{\phi^2} \;=\;\frac{3 - \sqrt{5}}{2}$