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Sagot :
Certainly! To determine the value of [tex]\(\alpha^3 + \beta^3 + \gamma^3\)[/tex] given that [tex]\(\alpha\)[/tex], [tex]\(\beta\)[/tex], and [tex]\(\gamma\)[/tex] are the zeros of the polynomial [tex]\(x^3 - 5x^2 + 6x - 1\)[/tex], let's follow a step-by-step approach.
### Step 1: Identify the Polynomial Coefficients and Use Vieta's Formulas
The polynomial we are working with is [tex]\(x^3 - 5x^2 + 6x - 1\)[/tex]. By Vieta's formulas, we know that for a polynomial [tex]\(x^3 + px^2 + qx + r\)[/tex], the sum and products of its roots can be expressed in terms of the coefficients:
- The sum of the roots, [tex]\(\alpha + \beta + \gamma\)[/tex], is [tex]\(-(-5) = 5\)[/tex].
- The sum of the product of the roots taken two at a time, [tex]\(\alpha\beta + \beta\gamma + \gamma\alpha\)[/tex], is [tex]\(6\)[/tex].
- The product of the roots, [tex]\(\alpha\beta\gamma\)[/tex], is [tex]\(-(-1) = 1\)[/tex].
So,
[tex]\[ \alpha + \beta + \gamma = 5 \][/tex]
[tex]\[ \alpha\beta + \beta\gamma + \gamma\alpha = 6 \][/tex]
[tex]\[ \alpha\beta\gamma = 1 \][/tex]
### Step 2: Compute the Sum of the Squares of the Roots
Using the identity for the sum of squares of the roots:
[tex]\[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \][/tex]
Substitute the values:
[tex]\[ \alpha^2 + \beta^2 + \gamma^2 = (5)^2 - 2(6) \][/tex]
[tex]\[ \alpha^2 + \beta^2 + \gamma^2 = 25 - 12 = 13 \][/tex]
### Step 3: Use the Identity for the Sum of the Cubes of the Roots
We use the identity for the sum of the cubes of the roots:
[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]\[ \alpha^3 + \beta^3 + \gamma^3 = (5)(13 - 6) + 3(1) \][/tex]
[tex]\[ \alpha^3 + \beta^3 + \gamma^3 = 5 \times 7 + 3 \][/tex]
[tex]\[ \alpha^3 + \beta^3 + \gamma^3 = 35 + 3 = 38 \][/tex]
### Conclusion
Hence, the value of [tex]\(\alpha^3 + \beta^3 + \gamma^3\)[/tex] is:
[tex]\[ \boxed{38} \][/tex]
### Step 1: Identify the Polynomial Coefficients and Use Vieta's Formulas
The polynomial we are working with is [tex]\(x^3 - 5x^2 + 6x - 1\)[/tex]. By Vieta's formulas, we know that for a polynomial [tex]\(x^3 + px^2 + qx + r\)[/tex], the sum and products of its roots can be expressed in terms of the coefficients:
- The sum of the roots, [tex]\(\alpha + \beta + \gamma\)[/tex], is [tex]\(-(-5) = 5\)[/tex].
- The sum of the product of the roots taken two at a time, [tex]\(\alpha\beta + \beta\gamma + \gamma\alpha\)[/tex], is [tex]\(6\)[/tex].
- The product of the roots, [tex]\(\alpha\beta\gamma\)[/tex], is [tex]\(-(-1) = 1\)[/tex].
So,
[tex]\[ \alpha + \beta + \gamma = 5 \][/tex]
[tex]\[ \alpha\beta + \beta\gamma + \gamma\alpha = 6 \][/tex]
[tex]\[ \alpha\beta\gamma = 1 \][/tex]
### Step 2: Compute the Sum of the Squares of the Roots
Using the identity for the sum of squares of the roots:
[tex]\[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \][/tex]
Substitute the values:
[tex]\[ \alpha^2 + \beta^2 + \gamma^2 = (5)^2 - 2(6) \][/tex]
[tex]\[ \alpha^2 + \beta^2 + \gamma^2 = 25 - 12 = 13 \][/tex]
### Step 3: Use the Identity for the Sum of the Cubes of the Roots
We use the identity for the sum of the cubes of the roots:
[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]\[ \alpha^3 + \beta^3 + \gamma^3 = (5)(13 - 6) + 3(1) \][/tex]
[tex]\[ \alpha^3 + \beta^3 + \gamma^3 = 5 \times 7 + 3 \][/tex]
[tex]\[ \alpha^3 + \beta^3 + \gamma^3 = 35 + 3 = 38 \][/tex]
### Conclusion
Hence, the value of [tex]\(\alpha^3 + \beta^3 + \gamma^3\)[/tex] is:
[tex]\[ \boxed{38} \][/tex]
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