Given that, $ x = (4096)^{7+4 \sqrt{3}} $
$ \Rightarrow x^{\frac{1}{7+4 \sqrt{3}}} = 4096 $
Do the rationalization, we get
$ \Rightarrow x^{ \left( \frac{1}{7+4 \sqrt{3}} \times \frac{7-4\sqrt{3}}{7-4\sqrt{3}} \right)} = 4096 $
$ \Rightarrow x^{ \left( \frac{7-4\sqrt{3}}{49-48} \right)} = 2^{12} \quad [ \because a^{2} – b^{2} = (a+b)(a-b)] $
$ \Rightarrow x^{7-4 \sqrt{3}} = (2^{6})^{2} $
$ \Rightarrow x^{7-4 \sqrt{3}} = (64)^{2} $
$ \Rightarrow x^{\frac{1}{2} (7-4 \sqrt{3})} = 64 $
$ \Rightarrow 64 = x^{\left( \frac{7-4 \sqrt{3}}{2}\right)} $
$ \Rightarrow 64 = x^{ \left(\frac{7}{2} - 2 \sqrt{3} \right)} $
$ \Rightarrow 64 = x^{\frac{7}{2}} \cdot x^{-2 \sqrt{3}} $
$ \Rightarrow \boxed{64 = \frac{x^{\frac{7}{2}}}{x^{2 \sqrt{3}}}} \quad \left[ \because x^{-a} = \dfrac{1}{x^{a}}\right] $
Correct Answer$: \text{D}$