If `lambda, mu` are real numbers such that , `x^(3)-lambdax^(2)+mux-6=0` has its real roots and positive, then the minimum value of `mu,` is

Options

A. a) `3(6)^(1//3)`
B. b) `3(6)^(2//3)`
C. c) `(6)^(1//3)`
D. d) `(6)^(2//3)`

Answer

To solve the problem, we need to find the minimum value of \( \mu \) such that the cubic equation

\[ x^3 - \lambda x^2 + \mu x - 6 = 0 \]

has three positive real roots. Let's denote the roots of the equation as \( \alpha, \beta, \gamma \).

### Step 1: Relate the coefficients to the roots
Using Vieta's formulas, we know:

1. The sum of the roots:
\[ \alpha + \beta + \gamma = \lambda \]

2. The sum of the product of the roots taken two at a time:
\[ \alpha\beta + \beta\gamma + \gamma\alpha = \mu \]

3. The product of the roots:
\[ \alpha\beta\gamma = 6 \]

### Step 2: Apply the AM-GM inequality
To find the minimum value of \( \mu \), we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality. According to AM-GM:

\[ \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{3} \geq \sqrt[3]{\alpha\beta \cdot \beta\gamma \cdot \gamma\alpha} \]

### Step 3: Express the right-hand side
The right-hand side can be expressed in terms of the product of the roots:

\[ \alpha\beta \cdot \beta\gamma \cdot \gamma\alpha = (\alpha\beta\gamma)^2 = (6)^2 = 36 \]

Thus, we have:

\[ \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{3} \geq \sqrt[3]{36} \]

### Step 4: Substitute for \( \mu \)
Substituting \( \mu \) into the inequality gives:

\[ \frac{\mu}{3} \geq \sqrt[3]{36} \]

### Step 5: Solve for \( \mu \)
Multiplying both sides by 3, we get:

\[ \mu \geq 3 \sqrt[3]{36} \]

### Step 6: Simplify \( \sqrt[3]{36} \)
We can simplify \( \sqrt[3]{36} \):

\[ \sqrt[3]{36} = \sqrt[3]{6^2} = 6^{2/3} \]

Thus, we have:

\[ \mu \geq 3 \cdot 6^{2/3} \]

### Conclusion
The minimum value of \( \mu \) is:

\[ \mu_{\text{min}} = 3 \cdot 6^{2/3} \]

### Final Answer
The minimum value of \( \mu \) is \( 3 \cdot 6^{2/3} \).