Simplify $\left(\sqrt[6]{27} - \sqrt{6 \frac{3}{4} }\right)^2$
$\textbf{(A)}\ \frac{3}{4} \qquad \textbf{(B)}\ \frac{\sqrt 3}{2} \qquad \textbf{(C)}\ \frac{3\sqrt 3}{4}\qquad \textbf{(D)}\ \frac{3}{2}\qquad \textbf{(E)}\ \frac{3\sqrt 3}{2}$
We have $\sqrt[6]{27} = (3^{3})^{\frac{1}{6}} = 3^{\frac{1}{2}} = \sqrt{3}$ and $\sqrt{6 \frac{3}{4}} = \sqrt{\frac{27}{4}} = \frac{3 \sqrt{3}}{2}$, so the answer is $(\sqrt{3} - \frac{3 \sqrt{3}}{2})^{2} = (-\frac{\sqrt{3}}{2})^{2} = \frac{3}{4}$, which is $\boxed{A}$.