If $$z^2 + z + 1 = 0$$, where $$z$$ is a complex number, then the value of $$$\left(z + \frac{1}{z}\right)^2 + \left(z^2 + \frac{1}{z^2}\right)^2 + \left(z^3 + \frac{1}{z^3}\right)^2 + \cdots + \left(z^6 + \frac{1}{z^6}\right)^2$$$ is

Given $$z^2 + z + 1 = 0$$, the roots are the non-real cube roots of unity. Hence

$$z^3 = 1 \quad\text{and}\quad z \neq 1, \qquad\text{with}\qquad 1/z = z^2.$$

For every positive integer $$k$$ we have $$\frac{1}{z^k} = (1/z)^k = (z^2)^k = z^{2k}.$$ Therefore

$$\left(z^k + \frac{1}{z^k}\right)^2 = \left(z^{\,k} + z^{\,2k}\right)^2.$$

Because $$z^3 = 1$$, any power of $$z$$ depends only on the remainder of the exponent when divided by 3:

$$z^{3m} = 1,\; z^{3m+1} = z,\; z^{3m+2} = z^2.$$

Compute each term for $$k = 1$$ to $$6$$.

$$z^{1} = z,\; z^{2\cdot1} = z^{2}.$$
Sum $$z + z^{2} = -1$$ (from $$z^2 + z + 1 = 0$$).
Square: $$(-1)^2 = 1.$$

$$z^{2} = z^{2},\; z^{4} = z^{1} = z.$$
Sum $$z^{2} + z = -1.$$. Square: $$1.$$

$$z^{3} = 1,\; z^{6} = (z^{3})^{2} = 1.$$
Sum $$1 + 1 = 2.$$ Square: $$4.$$

$$z^{4} = z^{1} = z,\; z^{8} = z^{2}.$$
Sum $$z + z^{2} = -1.$$. Square: $$1.$$

$$z^{5} = z^{2},\; z^{10} = z^{1} = z.$$
Sum $$z^{2} + z = -1.$$. Square: $$1.$$

$$z^{6} = 1,\; z^{12} = (z^{3})^{4} = 1.$$
Sum $$1 + 1 = 2.$$ Square: $$4.$$

Add all six results:

$$1 + 1 + 4 + 1 + 1 + 4 = 12.$$

Therefore, the required value is $$12$$.

Option D which is: 12