$(x^4+\frac{1}{x^4})^2=x^8+\frac{1}{x^8}+2(\frac{1}{x^4})(x^4)=47+2$ , therefore $x^4+\frac{1}{x^4}=7$
$(x^2+\frac{1}{x^2})^2=x^4+\frac{1}{x^4}+2=9$ , therefore $x^2+\frac{1}{x^2}=3$
$(x+\frac{1}{x})^2=x^2+\frac{1}{x^2}+2=5$ ,therefore $x+\frac{1}{x}=\sqrt5$
Now cube on both sides.
$(x+\frac{1}{x})^3=x^3+\frac{1}{x^3}+3(x+\frac{1}{x})$ this would give us $(x^3+\frac{1}{x^3})=5\sqrt5-3\sqrt5=2\sqrt5$
again cube on both sides.
$(x^3+\frac{1}{x^3})^3=x^9+\frac{1}{x^9}+3(x^3+\frac{1}{x^3})$ this will give us $(x^9+\frac{1}{x^9})=40\sqrt5-6\sqrt5=34\sqrt5$
Hence Option $(A)\space 34\sqrt5$ is correct.