If `2x^(3)+ax^(2)+bx+4=0` (a and b are positive real numbers) has three real roots.
The minimum value of `b^(3)` is a. 108 b. 216 c. 432 d. 864

Options

A. 432
B. 864
C. 1728
D. none of these

Explanation

To solve the problem, we need to analyze the cubic equation given:

\[ 2x^3 + ax^2 + bx + 4 = 0 \]

where \( a \) and \( b \) are positive real numbers, and we need to find the minimum value of \( b^3 \) such that the equation has three real roots.

### Step 1: Identify the conditions for three real roots

For the cubic equation to have three distinct real roots, the discriminant must be greater than zero. However, we can also use Vieta's formulas to express the relationships between the coefficients and the roots.

Let the roots be \( \alpha, \beta, \gamma \). According to Vieta's formulas:

1. The sum of the roots:
\[ \alpha + \beta + \gamma = -\frac{a}{2} \]

2. The sum of the products of the roots taken two at a time:
\[ \alpha\beta + \beta\gamma + \gamma\alpha = \frac{b}{2} \]

3. The product of the roots:
\[ \alpha\beta\gamma = -\frac{4}{2} = -2 \]

### Step 2: Apply the AM-GM inequality

Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality on the roots, we have:

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

Substituting the known values:

\[ \frac{\frac{b}{2}}{3} \geq \sqrt[3]{(-2)^2} \]

This simplifies to:

\[ \frac{b}{6} \geq \sqrt[3]{4} \]

### Step 3: Solve for \( b \)

From the inequality, we can solve for \( b \):

\[ b \geq 6 \cdot \sqrt[3]{4} \]

### Step 4: Cube both sides

Cubing both sides gives:

\[ b^3 \geq (6 \cdot \sqrt[3]{4})^3 \]

Calculating the right side:

\[ b^3 \geq 6^3 \cdot 4 = 216 \cdot 4 = 864 \]

### Conclusion

Thus, the minimum value of \( b^3 \) is:

\[ \boxed{864} \]