Question 4.

It is given that $x^8 + \left(\frac{1}{x}\right)^8 = 47$, which can be written as:

=> $\left(x^4\right)^2 + \left(\frac{1}{x^4}\right)^2 = 47$

=> $\left(x^4 + \frac{1}{x^4}\right)^2 - 2 \cdot x^4 \cdot \frac{1}{x^4} = 47$

=> $\left(x^4 + \frac{1}{x^4}\right)^2 = 49$

=> $x^4 + \frac{1}{x^4} = 7$

Similarly, $x^4 + \frac{1}{x^4} = 7$ can be expressed as:

=> $\left(x^2\right)^2 + \left(\frac{1}{x^2}\right)^2 = 7$

=> $\left(x^2 + \frac{1}{x^2}\right)^2 - 2 \cdot x^2 \cdot \frac{1}{x^2} = 7$

=> $\left(x^2 + \frac{1}{x^2}\right)^2 = 9$

=> $x^2 + \frac{1}{x^2} = 3$

By the same logic, we get $x + \frac{1}{x} = \sqrt{5}$

Now, $x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3 \cdot x \cdot \frac{1}{x}\left(x + \frac{1}{x}\right)$

=> $x^3 + \frac{1}{x^3} = \left(\sqrt{5}\right)^3 - 3\sqrt{5} = 2\sqrt{5}$

By the same logic, we can say that

=> $x^9 + \frac{1}{x^9} = \left(x^3 + \frac{1}{x^3}\right)^3 - 3 \cdot x^3 \cdot \frac{1}{x^3}\left(x^3 + \frac{1}{x^3}\right)$

=> $x^9 + \frac{1}{x^9} = \left(2\sqrt{5}\right)^3 - 3\left(2\sqrt{5}\right)$

=> $x^9 + \frac{1}{x^9} = 40\sqrt{5} - 6\sqrt{5} = 34\sqrt{5}$