Suppose the that quadratic equations `ax^(2)+bx+c=0 and bx^(2)+cx+a=0` have a common root. Then show that `a^(3)+b^(3)+c^(3)=3abc`.

Options

A. 3
B. 2
C. 1
D. 4

Answer

To solve the problem, we need to show that if the quadratic equations \( ax^2 + bx + c = 0 \) and \( bx^2 + cx + a = 0 \) have a common root, then \( a^3 + b^3 + c^3 = 3abc \).

### Step-by-Step Solution:

1. **Let the common root be \( \alpha \)**:
Since both equations have a common root, we can denote this common root as \( \alpha \).

2. **Substituting \( \alpha \) into the first equation**:
Substitute \( \alpha \) into the first equation:
\[ a\alpha^2 + b\alpha + c = 0 \quad \text{(1)} \]

3. **Substituting \( \alpha \) into the second equation**:
Substitute \( \alpha \) into the second equation:
\[ b\alpha^2 + c\alpha + a = 0 \quad \text{(2)} \]

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

5. **Substituting \( c \) into equation (2)**:
Substitute the expression for \( c \) from equation (3) into equation (2):
\[ b\alpha^2 - a\alpha^2 - b\alpha + a = 0 \]
Simplifying this gives:
\[ (b - a)\alpha^2 + (a - b)\alpha + a = 0 \]

6. **Rearranging terms**:
Rearranging the above equation:
\[ (b - a)\alpha^2 + (a - b)\alpha + a = 0 \]
This can be simplified to:
\[ (b - a)(\alpha^2 - \alpha) + a = 0 \]

7. **Considering cases**:
- If \( b \neq a \), we can divide through by \( b - a \):
\[ \alpha^2 - \alpha = -\frac{a}{b - a} \]
- If \( b = a \), then both equations become \( ax^2 + bx + c = 0 \) and \( ax^2 + cx + a = 0 \), which implies \( c = a \).

8. **Using the identity for cubes**:
We can use the identity:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
Since \( a = b \) or \( c = a \), we can show that \( a + b + c = 0 \) or \( a^2 + b^2 + c^2 - ab - ac - bc = 0 \).

9. **Conclusion**:
Therefore, we conclude that:
\[ a^3 + b^3 + c^3 = 3abc \]