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Sagot :
Certainly! Let's solve the problem step-by-step.
We are given a polynomial [tex]\( f(x) = x^3 - 5x^2 + 7x - 15 \)[/tex] and asked to find the value of [tex]\((\alpha \beta)^{-1} + (\beta \gamma)^{-1} + (\gamma \alpha)^{-1}\)[/tex], where [tex]\(\alpha, \beta, \gamma\)[/tex] are the roots of the polynomial.
First, let's recall Vieta's formulas, which relate the coefficients of the polynomial to sums and products of its roots. For a cubic polynomial of the form [tex]\( f(x) = x^3 + ax^2 + bx + c \)[/tex], the relationships are:
1. [tex]\(\alpha + \beta + \gamma = -\frac{\text{coefficient of } x^2}{\text{leading coefficient}}\)[/tex]
2. [tex]\(\alpha\beta + \beta\gamma + \gamma\alpha = \frac{\text{coefficient of } x}{\text{leading coefficient}}\)[/tex]
3. [tex]\(\alpha\beta\gamma = -\frac{\text{constant term}}{\text{leading coefficient}}\)[/tex]
For our polynomial [tex]\( f(x) = x^3 - 5x^2 + 7x - 15 \)[/tex], the leading coefficient (coefficient of [tex]\( x^3 \)[/tex]) is 1.
According to Vieta's formulas:
1. [tex]\(\alpha + \beta + \gamma = -\left(-5\right) = 5\)[/tex]
2. [tex]\(\alpha\beta + \beta\gamma + \gamma\alpha = 7\)[/tex]
3. [tex]\(\alpha\beta\gamma = -(-15) = 15\)[/tex]
We need to find the value of [tex]\((\alpha \beta)^{-1} + (\beta \gamma)^{-1} + (\gamma \alpha)^{-1}\)[/tex].
Notice that:
[tex]\[ (\alpha \beta)^{-1} + (\beta \gamma)^{-1} + (\gamma \alpha)^{-1} = \frac{1}{\alpha \beta} + \frac{1}{\beta \gamma} + \frac{1}{\gamma \alpha} \][/tex]
To simplify these fractions, we find a common denominator, which is:
[tex]\[ \alpha \beta \cdot \beta \gamma \cdot \gamma \alpha = (\alpha \beta \gamma)^2 \][/tex]
Now, we rewrite the expression:
[tex]\[ \frac{1}{\alpha \beta} + \frac{1}{\beta \gamma} + \frac{1}{\gamma \alpha} = \frac{\gamma \alpha \beta + \alpha \beta \gamma + \beta \gamma \alpha}{(\alpha \beta \gamma)^2} \][/tex]
Since addition in the numerator is simply the sum of all product pairs of roots, because of symmetry:
[tex]\[ = \frac{3\alpha \beta \gamma}{(\alpha \beta \gamma)^2} = \frac{3 \alpha \beta \gamma}{(\alpha \beta \gamma)^2} \][/tex]
We know from Vieta's formulas that:
[tex]\[ \alpha \beta \gamma = 15 \][/tex]
Substitute this into the expression:
[tex]\[ = \frac{3 \times 15}{(15)^2} = \frac{45}{225} = \frac{1}{5} \][/tex]
Therefore, the value of [tex]\((\alpha \beta)^{-1} + (\beta \gamma)^{-1} + (\gamma \alpha)^{-1}\)[/tex] is:
[tex]\[ \boxed{\frac{1}{5}} \][/tex]
We are given a polynomial [tex]\( f(x) = x^3 - 5x^2 + 7x - 15 \)[/tex] and asked to find the value of [tex]\((\alpha \beta)^{-1} + (\beta \gamma)^{-1} + (\gamma \alpha)^{-1}\)[/tex], where [tex]\(\alpha, \beta, \gamma\)[/tex] are the roots of the polynomial.
First, let's recall Vieta's formulas, which relate the coefficients of the polynomial to sums and products of its roots. For a cubic polynomial of the form [tex]\( f(x) = x^3 + ax^2 + bx + c \)[/tex], the relationships are:
1. [tex]\(\alpha + \beta + \gamma = -\frac{\text{coefficient of } x^2}{\text{leading coefficient}}\)[/tex]
2. [tex]\(\alpha\beta + \beta\gamma + \gamma\alpha = \frac{\text{coefficient of } x}{\text{leading coefficient}}\)[/tex]
3. [tex]\(\alpha\beta\gamma = -\frac{\text{constant term}}{\text{leading coefficient}}\)[/tex]
For our polynomial [tex]\( f(x) = x^3 - 5x^2 + 7x - 15 \)[/tex], the leading coefficient (coefficient of [tex]\( x^3 \)[/tex]) is 1.
According to Vieta's formulas:
1. [tex]\(\alpha + \beta + \gamma = -\left(-5\right) = 5\)[/tex]
2. [tex]\(\alpha\beta + \beta\gamma + \gamma\alpha = 7\)[/tex]
3. [tex]\(\alpha\beta\gamma = -(-15) = 15\)[/tex]
We need to find the value of [tex]\((\alpha \beta)^{-1} + (\beta \gamma)^{-1} + (\gamma \alpha)^{-1}\)[/tex].
Notice that:
[tex]\[ (\alpha \beta)^{-1} + (\beta \gamma)^{-1} + (\gamma \alpha)^{-1} = \frac{1}{\alpha \beta} + \frac{1}{\beta \gamma} + \frac{1}{\gamma \alpha} \][/tex]
To simplify these fractions, we find a common denominator, which is:
[tex]\[ \alpha \beta \cdot \beta \gamma \cdot \gamma \alpha = (\alpha \beta \gamma)^2 \][/tex]
Now, we rewrite the expression:
[tex]\[ \frac{1}{\alpha \beta} + \frac{1}{\beta \gamma} + \frac{1}{\gamma \alpha} = \frac{\gamma \alpha \beta + \alpha \beta \gamma + \beta \gamma \alpha}{(\alpha \beta \gamma)^2} \][/tex]
Since addition in the numerator is simply the sum of all product pairs of roots, because of symmetry:
[tex]\[ = \frac{3\alpha \beta \gamma}{(\alpha \beta \gamma)^2} = \frac{3 \alpha \beta \gamma}{(\alpha \beta \gamma)^2} \][/tex]
We know from Vieta's formulas that:
[tex]\[ \alpha \beta \gamma = 15 \][/tex]
Substitute this into the expression:
[tex]\[ = \frac{3 \times 15}{(15)^2} = \frac{45}{225} = \frac{1}{5} \][/tex]
Therefore, the value of [tex]\((\alpha \beta)^{-1} + (\beta \gamma)^{-1} + (\gamma \alpha)^{-1}\)[/tex] is:
[tex]\[ \boxed{\frac{1}{5}} \][/tex]
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