If $$m$$ is chosen in the quadratic equation $$(m^2 + 1)x^2 - 3x + (m^2 + 1)^2 = 0$$ such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is:

We start with the quadratic equation

$$ (m^{2}+1)x^{2}-3x+(m^{2}+1)^{2}=0 $$

Let its two real roots be $$\alpha$$ and $$\beta$$. For a general quadratic $$ax^{2}+bx+c=0$$ we know the standard relationships

$$ \alpha+\beta=\frac{-\,b}{a}, \qquad \alpha\beta=\frac{c}{a}. $$

Here we can read off

$$ a=m^{2}+1,\; b=-3,\; c=(m^{2}+1)^{2}. $$

So the sum of the roots is

$$ \alpha+\beta=\frac{-(-3)}{m^{2}+1}=\frac{3}{m^{2}+1}. $$

Because $$m^{2}+1>0$$ for every real $$m$$, the quantity $$3/(m^{2}+1)$$ is positive and will be largest when the denominator $$m^{2}+1$$ is as small as possible. The least value of $$m^{2}+1$$ is obtained when $$m=0$$, giving $$m^{2}+1=1$$. Therefore the sum of the roots is maximized when

$$ m=0\quad\text{and}\quad \alpha+\beta=3. $$

With $$m=0$$, the given quadratic reduces to

$$ x^{2}-3x+1=0. $$

We now find the actual roots. Using the quadratic formula for $$x^{2}-3x+1=0$$, we have

$$ x=\frac{3\pm\sqrt{(-3)^{2}-4\cdot1\cdot1}}{2\cdot1} =\frac{3\pm\sqrt{9-4}}{2} =\frac{3\pm\sqrt{5}}{2}. $$

Thus

$$ \alpha=\frac{3+\sqrt{5}}{2},\qquad \beta =\frac{3-\sqrt{5}}{2}. $$

We are asked for the absolute difference of the cubes of the roots, that is,

$$ \left|\alpha^{3}-\beta^{3}\right|. $$

To avoid long numerical cubing, we use the algebraic identity

$$ \alpha^{3}-\beta^{3}=(\alpha-\beta)\,(\alpha^{2}+\alpha\beta+\beta^{2}). $$

First we find $$\alpha-\beta$$:

$$ \alpha-\beta=\frac{3+\sqrt{5}}{2}-\frac{3-\sqrt{5}}{2} =\sqrt{5}. $$

Next we evaluate $$\alpha^{2}+\alpha\beta+\beta^{2}$$. We already have the convenient symmetric sums

$$ \alpha+\beta=3,\qquad \alpha\beta=1. $$

Now,

$$ (\alpha+\beta)^{2} = \alpha^{2}+2\alpha\beta+\beta^{2}. $$

Therefore,

$$ \alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2\alpha\beta =3^{2}-2\cdot1=9-2=7. $$

Adding $$\alpha\beta=1$$ to this gives

$$ \alpha^{2}+\alpha\beta+\beta^{2}=7+1=8. $$

Putting everything into the identity, we get

$$ \alpha^{3}-\beta^{3} =(\alpha-\beta)\,(\alpha^{2}+\alpha\beta+\beta^{2}) =(\sqrt{5})(8)=8\sqrt{5}. $$

The value is already positive, so its absolute value stays the same:

$$ \left|\alpha^{3}-\beta^{3}\right|=8\sqrt{5}. $$

Hence, the correct answer is Option D.