To find the value of $\alpha^2 + \beta^2 + \gamma^2$ for the given cubic equation $x^3 + 4x^2 + 6x + 1 = 0$, we can use the properties of the roots of polynomial equations. Specifically, we can utilize Vieta's formulas to relate the roots to the coefficients of the polynomial.

For a cubic equation $x^3 + ax^2 + bx + c = 0$ with roots $\alpha, \beta,$ and $\gamma$, Vieta's formulas state:

1. $\alpha + \beta + \gamma = -a$
2. $\alpha\beta + \beta\gamma + \gamma\alpha = b$
3. $\alpha\beta\gamma = -c$

In our case, the equation is $x^3 + 4x^2 + 6x + 1 = 0$, so:

• $a = 4$
• $b = 6$
• $c = 1$

Using Vieta's formulas:

1. $\alpha + \beta + \gamma = -4$
2. $\alpha\beta + \beta\gamma + \gamma\alpha = 6$
3. $\alpha\beta\gamma = -1$

To find $\alpha^2 + \beta^2 + \gamma^2$, we use the identity: $\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)$

Substitute the values from Vieta's formulas: $\alpha^2 + \beta^2 + \gamma^2 = (-4)^2 - 2 \cdot 6$ $\alpha^2 + \beta^2 + \gamma^2 = 16 - 12$ $\alpha^2 + \beta^2 + \gamma^2 = 4$

So, the value of $\alpha^2 + \beta^2 + \gamma^2$ is $4$.

Would you like more details or have any questions?

Here are some related questions you might find useful to explore:

1. How can Vieta's formulas be derived?
2. What is the significance of the roots in a polynomial equation?
3. How can complex roots affect the sum of squares calculation?
4. What are some applications of polynomial roots in real-world problems?
5. How do you solve cubic equations with non-real roots?

Tip: When dealing with polynomial equations, always consider using Vieta's formulas to quickly relate the roots to the coefficients.