We are given the equation:

$w^8 + \frac{1}{1 - 4\sqrt{5}} + \frac{1}{1 + 4\sqrt{5}} + \frac{-1 - \sqrt{5}}{2} = 0$

The problem asks for the sum of the roots of this equation, denoted by $v$, and the value of $v^2$.

Step 1: Analyze the equation

The roots of the equation $w_1, w_2, \ldots, w_8$ are provided as solutions to this polynomial equation. However, we first simplify the constants on the right-hand side of the equation.

Simplify each term:

1. Second term: $\frac{1}{1 - 4\sqrt{5}}$
2. Third term: $\frac{1}{1 + 4\sqrt{5}}$

We can sum these two terms:

$\frac{1}{1 - 4\sqrt{5}} + \frac{1}{1 + 4\sqrt{5}} = \frac{(1 + 4\sqrt{5}) + (1 - 4\sqrt{5})}{(1 - 4\sqrt{5})(1 + 4\sqrt{5})}$ $= \frac{1 + 4\sqrt{5} + 1 - 4\sqrt{5}}{1^2 - (4\sqrt{5})^2} = \frac{2}{1 - 80} = \frac{2}{-79}$

Thus, these two terms simplify to $\frac{-2}{79}$.

Fourth term:

$\frac{-1 - \sqrt{5}}{2}$

Step 2: Substituting into the equation

Now substitute the simplified values back into the original equation:

$w^8 + \frac{-2}{79} + \frac{-1 - \sqrt{5}}{2} = 0$

The equation becomes:

$w^8 = \frac{2}{79} + \frac{1 + \sqrt{5}}{2}$

Step 3: Sum of roots

The problem states that $w_1, w_2, \ldots, w_8$ are the roots of this equation. For an 8th-degree polynomial of the form $w^8 + \ldots = 0$, the sum of the roots (by Vieta’s formulas) is zero.

Thus, $v = 0$, and $v^2 = 0^2 = 0$.

Final Answer:

The correct answer is:

$\boxed{0}$

Would you like further clarification or details on any part of this solution?

Here are 5 related questions to expand your understanding:

1. What are Vieta’s formulas, and how do they apply to polynomial equations?
2. How can you simplify expressions involving square roots and radicals?
3. What is the general approach to solving 8th-degree polynomial equations?
4. How does the discriminant of a polynomial affect its number of real roots?
5. What methods can be used to find the roots of higher-degree polynomials?

Tip: Vieta’s formulas are useful for quickly finding relationships between the coefficients of polynomials and their roots without needing to fully solve the equation.