If x is a positive real number such that $$x^8 + \left(\frac{1}{x}\right)^8 = 47$$, then the value of $$x^9 + \left(\frac{1}{x}\right)^9$$ is

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 

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

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

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

The correct option is D