If `ax^2 + bx + c = 0 and bx^2 + cx+a= 0` have a common root and `a!=0` then `(a^3+b^3+c^3)/(abc)` is

Options

A. 1
B. 2
C. 3
D. none of these

Answer

To solve the problem, we need to find the value of \(\frac{a^3 + b^3 + c^3}{abc}\) given that the equations \(ax^2 + bx + c = 0\) and \(bx^2 + cx + a = 0\) have a common root, and \(a \neq 0\).

### Step-by-Step Solution:

1. **Identify the Common Root**:
Let \(\alpha\) be the common root of both equations. Thus, we have:
\[ a\alpha^2 + b\alpha + c = 0 \quad \text{(1)} \]
\[ b\alpha^2 + c\alpha + a = 0 \quad \text{(2)} \]

2. **Express \(\alpha^2\) and \(\alpha\)**:
From equation (1), we can express \(\alpha^2\):
\[ a\alpha^2 = -b\alpha - c \implies \alpha^2 = \frac{-b\alpha - c}{a} \quad \text{(3)} \]
From equation (2), we can express \(\alpha\):
\[ b\alpha^2 = -c\alpha - a \implies \alpha = \frac{-b\alpha^2 - a}{c} \quad \text{(4)} \]

3. **Substituting \(\alpha^2\) into \(\alpha\)**:
Substitute equation (3) into equation (4):
\[ \alpha = \frac{-b\left(\frac{-b\alpha - c}{a}\right) - a}{c} \]
Simplifying this gives:
\[ \alpha = \frac{b(b\alpha + c) - ac}{ac} = \frac{b^2\alpha + bc - ac}{c} \]

4. **Rearranging**:
Rearranging the equation gives:
\[ c\alpha = b^2\alpha + bc - ac \]
\[ (c - b^2)\alpha = bc - ac \]

5. **Finding \(\alpha\)**:
If \(c - b^2 \neq 0\), we can solve for \(\alpha\):
\[ \alpha = \frac{bc - ac}{c - b^2} \]

6. **Using the Result**:
Since both equations have a common root, we can use the condition derived from the equations:
\[ a^3 + b^3 + c^3 - 3abc = 0 \]
This implies:
\[ a^3 + b^3 + c^3 = 3abc \]

7. **Final Calculation**:
Now, substituting this result into the expression we need:
\[ \frac{a^3 + b^3 + c^3}{abc} = \frac{3abc}{abc} = 3 \]

### Conclusion:
Thus, the value of \(\frac{a^3 + b^3 + c^3}{abc}\) is \(\boxed{3}\).