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

Options

A. 108
B. 216
C. 432
D. 864

Explanation

To solve the problem, we need to analyze the cubic equation given and apply some properties of roots. The equation is:

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

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

### Step 1: Identify the roots
Let the roots of the equation be \( \alpha, \beta, \gamma \). By Vieta's formulas, we know:

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 Arithmetic Mean-Geometric Mean Inequality (AM-GM)
Since the equation has three real roots, we can apply the AM-GM inequality to the roots \( -\alpha, -\beta, -\gamma \):

\[ -\frac{\alpha + \beta + \gamma}{3} \geq \sqrt[3]{(-\alpha)(-\beta)(-\gamma)} \]

This simplifies to:

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

### Step 3: Substitute the values from Vieta's formulas
Substituting the values from Vieta's formulas into the AM-GM inequality:

\[ -\frac{-\frac{a}{2}}{3} \geq \sqrt[3]{-2} \]

This gives:

\[ \frac{a}{6} \geq \sqrt[3]{2} \]

### Step 4: Cube both sides
Cubing both sides to eliminate the cube root:

\[ \left(\frac{a}{6}\right)^3 \geq 2 \]

This simplifies to:

\[ \frac{a^3}{216} \geq 2 \]

### Step 5: Solve for \( a^3 \)
Multiplying both sides by 216:

\[ a^3 \geq 432 \]

### Conclusion
Thus, the minimum value of \( a^3 \) is \( 432 \). Therefore, the answer is:

**c. 432**