Answer :
Certainly! Let's solve the two parts of this problem step by step:
### Part (a)
Expand [tex]\((\alpha \beta + \beta \gamma + \gamma \alpha)^2\)[/tex]:
We will use the formula for squaring a sum of three terms:
[tex]\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \][/tex]
Here, let:
[tex]\(a = \alpha \beta\)[/tex]
[tex]\(b = \beta \gamma\)[/tex]
[tex]\(c = \gamma \alpha\)[/tex]
Then,
[tex]\[ (\alpha \beta + \beta \gamma + \gamma \alpha)^2 = (\alpha \beta)^2 + (\beta \gamma)^2 + (\gamma \alpha)^2 + 2(\alpha \beta)(\beta \gamma) + 2(\beta \gamma)(\gamma \alpha) + 2(\gamma \alpha)(\alpha \beta) \][/tex]
Now calculate each term:
[tex]\[ (\alpha \beta)^2 = \alpha^2 \beta^2 \][/tex]
[tex]\[ (\beta \gamma)^2 = \beta^2 \gamma^2 \][/tex]
[tex]\[ (\gamma \alpha)^2 = \gamma^2 \alpha^2 \][/tex]
[tex]\[ 2(\alpha \beta)(\beta \gamma) = 2 \alpha \beta^2 \gamma \][/tex]
[tex]\[ 2(\beta \gamma)(\gamma \alpha) = 2 \beta \gamma^2 \alpha \][/tex]
[tex]\[ 2(\gamma \alpha)(\alpha \beta) = 2 \gamma \alpha^2 \beta \][/tex]
So, when combined, we get:
[tex]\[ (\alpha \beta + \beta \gamma + \gamma \alpha)^2 = \alpha^2 \beta^2 + \beta^2 \gamma^2 + \gamma^2 \alpha^2 + 2 \alpha \beta^2 \gamma + 2 \beta \gamma^2 \alpha + 2 \gamma \alpha^2 \beta \][/tex]
### Part (b)
If [tex]\(\alpha, \beta, \gamma\)[/tex] are the roots of the equation [tex]\(x^3 - x^2 + 2x + 6 = 0\)[/tex], we need to find a cubic equation with roots [tex]\(\alpha^2, \beta^2, \gamma^2\)[/tex].
First, we recall Vieta's formulas for the roots of the polynomial [tex]\(x^3 - x^2 + 2x + 6 = 0\)[/tex]:
- Sum of the roots, [tex]\(\alpha + \beta + \gamma = 1\)[/tex]
- Sum of the product of the roots taken two at a time, [tex]\(\alpha \beta + \beta \gamma + \gamma \alpha = 2\)[/tex]
- Product of the roots, [tex]\(\alpha \beta \gamma = -6\)[/tex]
We want to find the polynomial with roots [tex]\(\alpha^2, \beta^2, \gamma^2\)[/tex]. Using the relations for new roots, we find:
1. The sum of the new roots ([tex]\(\alpha^2 + \beta^2 + \gamma^2\)[/tex]):
[tex]\[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha \beta + \beta \gamma + \gamma \alpha) = 1^2 - 2 \cdot 2 = 1 - 4 = -3 \][/tex]
2. The sum of the product of the new roots taken two at a time ([tex]\(\alpha^2 \beta^2 + \beta^2 \gamma^2 + \gamma^2 \alpha^2\)[/tex]):
[tex]\[ \alpha^2 \beta^2 + \beta^2 \gamma^2 + \gamma^2 \alpha^2 = (\alpha \beta + \beta \gamma + \gamma \alpha)^2 - 2(\alpha \beta \gamma)(\alpha + \beta + \gamma) \][/tex]
[tex]\[ = 2^2 - 2(-6)(1) = 4 + 12 = 16 \][/tex]
3. The product of the new roots ([tex]\(\alpha^2 \beta^2 \gamma^2\)[/tex]):
[tex]\[ (\alpha \beta \gamma)^2 = (-6)^2 = 36 \][/tex]
So, the polynomial with roots [tex]\(\alpha^2, \beta^2, \gamma^2\)[/tex] is:
[tex]\[ x^3 - (\text{sum of roots})x^2 + (\text{sum of products of roots taken two at a time})x - (\text{product of roots}) = 0 \][/tex]
[tex]\[ x^3 - (-3)x^2 + 16x - 36 = 0 \][/tex]
[tex]\[ x^3 + 3x^2 + 16x - 36 = 0 \][/tex]
Therefore, the cubic equation with roots [tex]\(\alpha^2, \beta^2, \gamma^2\)[/tex] is:
[tex]\[ \boxed{x^3 + 3x^2 + 16x - 36 = 0} \][/tex]
### Part (a)
Expand [tex]\((\alpha \beta + \beta \gamma + \gamma \alpha)^2\)[/tex]:
We will use the formula for squaring a sum of three terms:
[tex]\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \][/tex]
Here, let:
[tex]\(a = \alpha \beta\)[/tex]
[tex]\(b = \beta \gamma\)[/tex]
[tex]\(c = \gamma \alpha\)[/tex]
Then,
[tex]\[ (\alpha \beta + \beta \gamma + \gamma \alpha)^2 = (\alpha \beta)^2 + (\beta \gamma)^2 + (\gamma \alpha)^2 + 2(\alpha \beta)(\beta \gamma) + 2(\beta \gamma)(\gamma \alpha) + 2(\gamma \alpha)(\alpha \beta) \][/tex]
Now calculate each term:
[tex]\[ (\alpha \beta)^2 = \alpha^2 \beta^2 \][/tex]
[tex]\[ (\beta \gamma)^2 = \beta^2 \gamma^2 \][/tex]
[tex]\[ (\gamma \alpha)^2 = \gamma^2 \alpha^2 \][/tex]
[tex]\[ 2(\alpha \beta)(\beta \gamma) = 2 \alpha \beta^2 \gamma \][/tex]
[tex]\[ 2(\beta \gamma)(\gamma \alpha) = 2 \beta \gamma^2 \alpha \][/tex]
[tex]\[ 2(\gamma \alpha)(\alpha \beta) = 2 \gamma \alpha^2 \beta \][/tex]
So, when combined, we get:
[tex]\[ (\alpha \beta + \beta \gamma + \gamma \alpha)^2 = \alpha^2 \beta^2 + \beta^2 \gamma^2 + \gamma^2 \alpha^2 + 2 \alpha \beta^2 \gamma + 2 \beta \gamma^2 \alpha + 2 \gamma \alpha^2 \beta \][/tex]
### Part (b)
If [tex]\(\alpha, \beta, \gamma\)[/tex] are the roots of the equation [tex]\(x^3 - x^2 + 2x + 6 = 0\)[/tex], we need to find a cubic equation with roots [tex]\(\alpha^2, \beta^2, \gamma^2\)[/tex].
First, we recall Vieta's formulas for the roots of the polynomial [tex]\(x^3 - x^2 + 2x + 6 = 0\)[/tex]:
- Sum of the roots, [tex]\(\alpha + \beta + \gamma = 1\)[/tex]
- Sum of the product of the roots taken two at a time, [tex]\(\alpha \beta + \beta \gamma + \gamma \alpha = 2\)[/tex]
- Product of the roots, [tex]\(\alpha \beta \gamma = -6\)[/tex]
We want to find the polynomial with roots [tex]\(\alpha^2, \beta^2, \gamma^2\)[/tex]. Using the relations for new roots, we find:
1. The sum of the new roots ([tex]\(\alpha^2 + \beta^2 + \gamma^2\)[/tex]):
[tex]\[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha \beta + \beta \gamma + \gamma \alpha) = 1^2 - 2 \cdot 2 = 1 - 4 = -3 \][/tex]
2. The sum of the product of the new roots taken two at a time ([tex]\(\alpha^2 \beta^2 + \beta^2 \gamma^2 + \gamma^2 \alpha^2\)[/tex]):
[tex]\[ \alpha^2 \beta^2 + \beta^2 \gamma^2 + \gamma^2 \alpha^2 = (\alpha \beta + \beta \gamma + \gamma \alpha)^2 - 2(\alpha \beta \gamma)(\alpha + \beta + \gamma) \][/tex]
[tex]\[ = 2^2 - 2(-6)(1) = 4 + 12 = 16 \][/tex]
3. The product of the new roots ([tex]\(\alpha^2 \beta^2 \gamma^2\)[/tex]):
[tex]\[ (\alpha \beta \gamma)^2 = (-6)^2 = 36 \][/tex]
So, the polynomial with roots [tex]\(\alpha^2, \beta^2, \gamma^2\)[/tex] is:
[tex]\[ x^3 - (\text{sum of roots})x^2 + (\text{sum of products of roots taken two at a time})x - (\text{product of roots}) = 0 \][/tex]
[tex]\[ x^3 - (-3)x^2 + 16x - 36 = 0 \][/tex]
[tex]\[ x^3 + 3x^2 + 16x - 36 = 0 \][/tex]
Therefore, the cubic equation with roots [tex]\(\alpha^2, \beta^2, \gamma^2\)[/tex] is:
[tex]\[ \boxed{x^3 + 3x^2 + 16x - 36 = 0} \][/tex]
