Let $$\alpha$$ and $$\beta$$ be the roots of the equation $$x^2 + 2x + 2 = 0$$, then $$\alpha^{15} + \beta^{15}$$ is equal to:

We have the quadratic equation $$x^{2}+2x+2=0$$ whose roots are $$\alpha$$ and $$\beta$$.

For any quadratic equation $$x^{2}+px+q=0$$ with roots $$r_{1}$$ and $$r_{2}$$, we know from Vieta’s relations that $$r_{1}+r_{2}=-p$$ and $$r_{1}r_{2}=q$$. Comparing, we obtain:

$$\alpha+\beta=-2,\qquad \alpha\beta=2.$$

We wish to find $$\alpha^{15}+\beta^{15}$$. Because $$\alpha$$ and $$\beta$$ satisfy their own quadratic, we can express higher powers in terms of lower ones. Starting from the given equation, write it as

$$x^{2}=-2x-2.$$

Replacing $$x$$ by $$\alpha$$ and then by $$\beta$$ gives the identities

$$\alpha^{2}=-2\alpha-2,\qquad \beta^{2}=-2\beta-2.$$

Multiplying both sides by successive powers of the same root allows us to build a recurrence. In general, for $$n\ge 2$$,

$$\alpha^{n}= -2\alpha^{\,n-1}-2\alpha^{\,n-2},\qquad \beta^{n}= -2\beta^{\,n-1}-2\beta^{\,n-2}.$$

Adding these two relations term-by-term, define $$S_{n}=\alpha^{n}+\beta^{n}$$. We then have the recurrence relation

$$S_{n} = -2S_{n-1}-2S_{n-2}\qquad\text{for }n\ge 2.$$

To use it, compute the first two terms directly:

$$S_{0}=\alpha^{0}+\beta^{0}=1+1=2,$$

$$S_{1}=\alpha+\beta=-2.$$

Now apply the recurrence step by step:

$$S_{2} = -2S_{1}-2S_{0}= -2(-2)-2(2)=4-4=0,$$

$$S_{3} = -2S_{2}-2S_{1}= -2(0)-2(-2)=0+4=4,$$

$$S_{4} = -2S_{3}-2S_{2}= -2(4)-2(0)= -8,$$

$$S_{5} = -2S_{4}-2S_{3}= -2(-8)-2(4)=16-8=8,$$

$$S_{6} = -2S_{5}-2S_{4}= -2(8)-2(-8)= -16+16=0,$$

$$S_{7} = -2S_{6}-2S_{5}= -2(0)-2(8)= -16,$$

$$S_{8} = -2S_{7}-2S_{6}= -2(-16)-2(0)=32,$$

$$S_{9} = -2S_{8}-2S_{7}= -2(32)-2(-16)= -64+32= -32,$$

$$S_{10}= -2S_{9}-2S_{8}= -2(-32)-2(32)=64-64=0,$$

$$S_{11}= -2S_{10}-2S_{9}= -2(0)-2(-32)=64,$$

$$S_{12}= -2S_{11}-2S_{10}= -2(64)-2(0)= -128,$$

$$S_{13}= -2S_{12}-2S_{11}= -2(-128)-2(64)=256-128=128,$$

$$S_{14}= -2S_{13}-2S_{12}= -2(128)-2(-128)= -256+256=0,$$

$$S_{15}= -2S_{14}-2S_{13}= -2(0)-2(128)= -256.$$

Thus, $$\alpha^{15}+\beta^{15}=S_{15}=-256.$$

Hence, the correct answer is Option D.