Certainly! Let's analyze and solve the problem step-by-step. We're given the polynomial [tex]\( p(x) = x^3 - 7x^2 + 11x - 7 \)[/tex] with roots [tex]\(\alpha\)[/tex], [tex]\(\beta\)[/tex], and [tex]\(\gamma\)[/tex]. We need to determine two specific sums involving these roots.
### Step 1: Sum and Products of Roots Using Vieta's Formulas
1. Sum of Roots:
From Vieta's formulas for a cubic polynomial [tex]\(ax^3 + bx^2 + cx + d\)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -7 \)[/tex], [tex]\( c = 11 \)[/tex], and [tex]\( d = -7 \)[/tex]:
The sum of the roots is:
[tex]\[
\alpha + \beta + \gamma = -\frac{b}{a} = -\frac{-7}{1} = 7
\][/tex]
2. Sum of the Product of Roots taken two at a time:
[tex]\[
\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{11}{1} = 11
\][/tex]
3. Product of Roots:
[tex]\[
\alpha\beta\gamma = -\frac{d}{a} = -\frac{-7}{1} = 7
\][/tex]
### Step 2: Calculation of [tex]\(\alpha^2 + \beta^2 + \gamma^2\)[/tex]
Using the identity:
[tex]\[
\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)
\][/tex]
Substitute the known values:
[tex]\[
\alpha^2 + \beta^2 + \gamma^2 = 7^2 - 2 \cdot 11 = 49 - 22 = 27
\][/tex]
### Step 3: Calculation of [tex]\(\alpha^3 + \beta^3 + \gamma^3\)[/tex]
Using the identity:
[tex]\[
\alpha^3 + \beta^3 + \gamma^3 = (\alpha + \beta + \gamma)((\alpha^2 + \beta^2 + \gamma^2) - (\alpha\beta + \beta\gamma + \gamma\alpha)) + 3\alpha\beta\gamma
\][/tex]
Substitute the known values:
[tex]\[
\alpha^3 + \beta^3 + \gamma^3 = 7 \left(27 - 11\right) + 3 \cdot 7
\][/tex]
[tex]\[
\alpha^3 + \beta^3 + \gamma^3 = 7 \cdot 16 + 21 = 112 + 21 = 133
\][/tex]
### Summary:
i) The value of [tex]\(\alpha^2 + \beta^2 + \gamma^2\)[/tex] is [tex]\(27\)[/tex].
ii) The value of [tex]\(\alpha^3 + \beta^3 + \gamma^3\)[/tex] is [tex]\(133\)[/tex].
