If $$a^2 + b^2 + 64c^2 + 16c + 3 = 2(a + b)$$, then the value of $$4a^7 + b^7 + 8c^2$$ is:

From the given equation,

$$a^2 + b^2 + 64c^2 + 16c + 3 = 2(a + b)$$

$$\Rightarrow a^2 + b^2 + 64c^2 + 16c + 3 -2a-2b= 0$$

$$\Rightarrow a^2-2a+1 + b^2-2b+1 + (8c)^2 + 16c +1= 0$$

$$\Rightarrow (a-1)^2+ (b-1)^2 + (8c+1)^2= 0$$

It will be 0, if every terms individually equal to 0.

So, a=1, b=1, $$c=\dfrac{-1}{8}$$

Now, substituting the values in the equation $$4a^7 + b^7 + 8c^2$$

$$\Rightarrow 4(1)^7 + (1)^7 + 8(\dfrac{-1}{8})^2$$

$$\Rightarrow 4 + 1 + (\dfrac{1}{8})$$

$$\Rightarrow \dfrac{41}{8}=5\dfrac{1}{8}$$