1. Identify $z$ in trigonometric form

Given

$$z=\frac{1-i\sqrt3}{2}$$

we note that

$$\cos\frac{\pi}{3}=\frac12,\qquad \sin\frac{\pi}{3}=\frac{\sqrt3}{2}$$

so

$$z=\frac12-i\frac{\sqrt3}{2}=\cos\frac{\pi}{3}-i\sin\frac{\pi}{3}=e^{-i\pi/3}.$$

Hence,

$$z^k=e^{-ik\pi/3},\qquad \frac1{z^k}=e^{ik\pi/3}.$$

Therefore,

$$z^k+\frac1{z^k}=e^{-ik\pi/3}+e^{ik\pi/3}=2\cos\frac{k\pi}{3}.$$

2. Rewrite the sum

The given expression is

$$21+\sum_{k=1}^{21}\left(z^k+\frac1{z^k}\right)^3.$$

Using the above result,

$$\left(z^k+\frac1{z^k}\right)^3=\left(2\cos\frac{k\pi}{3}\right)^3=8\cos^3\frac{k\pi}{3}.$$

So the sum becomes

$$S=21+\sum_{k=1}^{21}8\cos^3\frac{k\pi}{3}.$$

3. Find the repeating values of $\cos\frac{k\pi}{3}$

As $k$ runs modulo $6$,

$$\cos\frac{k\pi}{3} = \left(\frac12,-\frac12,-1,-\frac12,\frac12,1\right)$$

for $k=1,2,3,4,5,6$ respectively.

Thus,

$$8\cos^3\frac{k\pi}{3}$$

takes values

$$1,-1,-8,-1,1,8.$$

Their sum over one full block of $6$ terms is

$$1+(-1)+(-8)+(-1)+1+8=0.$$

4. Use periodicity up to $21$ terms

Since $21=18+3$, we have $3$ complete blocks of $6$ terms contributing zero, and then the first $3$ terms remain:

$$1,-1,-8.$$

Their sum is

$$1-1-8=-8.$$

Therefore,

$$\sum_{k=1}^{21}8\cos^3\frac{k\pi}{3}=-8.$$

So,

$$S=21+(-8)=13.$$

5. Final answer

$$\boxed{13}$$

The derived answer matches the stored correct answer.