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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.

Learn how to find the value of α² + β² + γ² for the cubic equation x³ + 4x² + 6x + 1 = 0 using Vieta's Formulas. This problem explores polynomial roots and their relationship with coefficients, suitable for advanced high school students.

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