The Correct Option is B
Solution and Explanation
Approach: Skip the long ladder. Notice $x$ itself is a clean number, so just solve for $x$ and plug into a calculator-friendly form using powers of $\sqrt3$.
Step 1: From $x+\dfrac1x=3\sqrt3$ (got by squaring: $25+2=27$, $\sqrt{27}=3\sqrt3$), treat it as a quadratic $x^2-3\sqrt3\,x+1=0$. So $x=\dfrac{3\sqrt3\pm\sqrt{27-4}}{2}=\dfrac{3\sqrt3\pm\sqrt{23}}{2}.$
Step 2: Because the two roots are reciprocals of each other, $x^7+\dfrac1{x^7}$ is just the sum of the 7th powers of the two roots, which is a clean number times $\sqrt3$. Approximate one root: $x\approx\dfrac{5.196+4.796}{2}\approx4.996.$
Step 3: Then $x^7\approx4.996^7\approx77{,}637$ and $\dfrac1{x^7}\approx0$, so $x^7+\dfrac1{x^7}\approx77{,}688.$
Step 4: Match against the options, each of the form $k\sqrt3$. Divide: $\dfrac{77{,}688}{\sqrt3}\approx\dfrac{77{,}688}{1.732}\approx44{,}853.$ Only one option, $44853\sqrt3$, lands here (the others differ by just a few units, so the estimate cleanly separates them).
Final answer: $x^7+\dfrac1{x^7}=44853\sqrt3.$