Algebraic Identity Expansion

Let $k = x + \frac{1}{x}$.

Since $x^8 + \frac{1}{x^8} = 47$, we work backwards: $\left(x^4 + \frac{1}{x^4}\right)^2 = x^8 + \frac{1}{x^8} + 2 = 47 + 2 = 49 \implies x^4 + \frac{1}{x^4} = 7$ $\left(x^2 + \frac{1}{x^2}\right)^2 = x^4 + \frac{1}{x^4} + 2 = 7 + 2 = 9 \implies x^2 + \frac{1}{x^2} = 3$ $\left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2 = 3 + 2 = 5 \implies x + \frac{1}{x} = \sqrt{5}$

Now, calculate $x^3 + \frac{1}{x^3}$ and $x^6 + \frac{1}{x^6}$: $x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right) = (\sqrt{5})^3 - 3\sqrt{5} = 2\sqrt{5}$ $x^6 + \frac{1}{x^6} = \left(x^3 + \frac{1}{x^3}\right)^2 - 2 = (2\sqrt{5})^2 - 2 = 20 - 2 = 18$

Finally, calculate $x^9 + \frac{1}{x^9}$: $x^9 + \frac{1}{x^9} = \left(x^6 + \frac{1}{x^6}\right)\left(x^3 + \frac{1}{x^3}\right) - \left(x^3 + \frac{1}{x^3}\right)$ $= 18(2\sqrt{5}) - 2\sqrt{5} = 36\sqrt{5} - 2\sqrt{5} = 34\sqrt{5}$