Step 1: Plan with polar form.
Powers and divisions of complex numbers are easiest in polar form, where the size (modulus) and the angle (argument) are handled separately. We will turn each piece into size and angle.
Step 2: Polar form of the numerator base.
For $1-i$, the size is $\sqrt{1^2+(-1)^2}=\sqrt{2}$ and the angle is $-45^\circ$. So $1-i=\sqrt{2}\,\text{cis}(-45^\circ)$. Cubing it: size becomes $(\sqrt{2})^3=2\sqrt{2}$ and angle becomes $-135^\circ$.
Step 3: Polar form of the denominator base.
For $\sqrt{3}-i$, the size is $\sqrt{3+1}=2$ and the angle is $-30^\circ$. So $\sqrt{3}-i=2\,\text{cis}(-30^\circ)$. Squaring it: size becomes $4$ and angle becomes $-60^\circ$.
Step 4: Divide the two.
For division, divide sizes and subtract angles. Size $=\frac{2\sqrt{2}}{4}=\frac{\sqrt{2}}{2}$. Angle $=-135^\circ-(-60^\circ)=-75^\circ$. So $z=\frac{\sqrt2}{2}\,\text{cis}(-75^\circ)$.
Step 5: Use known values for $75^\circ$.
We know $\cos 75^\circ=\frac{\sqrt6-\sqrt2}{4}$ and $\sin 75^\circ=\frac{\sqrt6+\sqrt2}{4}$. So $z=\frac{\sqrt2}{2}\left(\cos 75^\circ - i\sin 75^\circ\right)$.
Step 6: Multiply out and simplify.
The real part is $\frac{\sqrt2}{2}\cdot\frac{\sqrt6-\sqrt2}{4}=\frac{\sqrt{12}-2}{8}=\frac{2\sqrt3-2}{8}=\frac{\sqrt3-1}{4}$. The imaginary part of $z$ is $-\frac{\sqrt2}{2}\cdot\frac{\sqrt6+\sqrt2}{4}=-\frac{\sqrt3+1}{4}$. So $z=\frac{\sqrt3-1}{4}-\frac{\sqrt3+1}{4}i$.
Step 7: Take the conjugate.
The conjugate flips the sign of the imaginary part, and writing it in the same grouped form as the options gives:
\[ \boxed{\dfrac{\sqrt{3}+1}{4}-\dfrac{\sqrt{3}-1}{4}i} \]