If `alpha` and `beta`, `alpha` and `gamma`, `alpha` and `delta` are the roots of the equations `ax^(2)+2bx+c=0`, `2bx^(2)+cx+a=0` and `cx^(2)+ax+2b=0` respectively where `a`, `b`, `c` are positive real numbers, then `alpha+alpha^(2)` is equal to

Options

A. `-1`
B. `1`
C. `0`
D. `abc`

Explanation

To solve the problem, we need to analyze the three equations given in the question and find the value of \( \alpha + \alpha^2 \).

### Step 1: Write down the equations
The roots of the equations are given as follows:
1. \( ax^2 + 2bx + c = 0 \) (roots: \( \alpha, \beta \))
2. \( 2bx^2 + cx + a = 0 \) (roots: \( \alpha, \gamma \))
3. \( cx^2 + ax + 2b = 0 \) (roots: \( \alpha, \delta \))

### Step 2: Substitute \( \alpha \) into each equation
Since \( \alpha \) is a root of all three equations, we can substitute \( \alpha \) into each equation:

1. From the first equation:
\[ a\alpha^2 + 2b\alpha + c = 0 \]

2. From the second equation:
\[ 2b\alpha^2 + c\alpha + a = 0 \]

3. From the third equation:
\[ c\alpha^2 + a\alpha + 2b = 0 \]

### Step 3: Add the three equations
Now, we add all three equations together:
\[ (a\alpha^2 + 2b\alpha + c) + (2b\alpha^2 + c\alpha + a) + (c\alpha^2 + a\alpha + 2b) = 0 \]

### Step 4: Combine like terms
Combining the terms gives us:
\[ (a + 2b + c)\alpha^2 + (2b + c + a)\alpha + (c + a + 2b) = 0 \]

### Step 5: Factor out common terms
We can factor out \( (a + 2b + c) \):
\[ (a + 2b + c)(\alpha^2 + \alpha + 1) = 0 \]

### Step 6: Analyze the factor
Since \( a, b, c \) are positive real numbers, \( a + 2b + c \) cannot be zero. Therefore, we must have:
\[ \alpha^2 + \alpha + 1 = 0 \]

### Step 7: Solve for \( \alpha + \alpha^2 \)
The equation \( \alpha^2 + \alpha + 1 = 0 \) can be rearranged to find \( \alpha + \alpha^2 \):
\[ \alpha + \alpha^2 = -1 \]

### Conclusion
Thus, the value of \( \alpha + \alpha^2 \) is:
\[ \boxed{-1} \]