Given the quadratic equation:

$x^2 - 5x - 1 = 0$

Let $\alpha$ and $\beta$ be the roots of the equation. We want to find the value of $\alpha^3 + \beta^3$.

Step 1: Use Vieta's Formulas

From Vieta's formulas, for the roots of the equation $ax^2 + bx + c = 0$, we have:

• $\alpha + \beta = -\frac{b}{a} = 5$
• $\alpha \beta = \frac{c}{a} = -1$

Step 2: Use the Identity for $\alpha^3 + \beta^3$

We use the identity:

$\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha \beta + \beta^2)$

We need to find $\alpha^2 + \beta^2$:

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 5^2 - 2(-1) = 25 + 2 = 27$

Now substitute $\alpha^2 + \beta^2$ into the identity:

$\alpha^3 + \beta^3 = ( \alpha + \beta ) \left(\alpha^2 - \alpha \beta + \beta^2\right) = (5) \left(27 - (-1)\right) = 5 \times 28 = 140$

Therefore, the value of $\alpha^3 + \beta^3$ is:

$\boxed{140}$

Would you like further details or have any other questions?

Here are some related questions that you might find interesting:

1. How do Vieta's formulas apply to cubic equations?
2. Can you find the sum of the squares of the roots of a quadratic equation?
3. What are other identities involving the roots of a polynomial?
4. How can you express higher powers of roots in terms of coefficients?
5. How would the solution change if the quadratic equation had complex roots?

Tip: When working with polynomial roots, leveraging identities and relationships like Vieta's formulas can simplify calculations significantly.