If two roots of `x ^(3) -ax ^(2) + bx -c =0` are equal in magnitude but opposite in sign. Then:

Options

A. `a+bc=0`
B. `a ^(2) =bc`
C. `ab =c`
D. `a-b+c=0`

Answer

To solve the problem, we need to analyze the given cubic equation \( x^3 - ax^2 + bx - c = 0 \) under the condition that two of its roots are equal in magnitude but opposite in sign.

### Step-by-Step Solution:

1. **Define the Roots**:
Let the roots of the equation be \( \alpha, \beta, \) and \( \gamma \). Given that two roots are equal in magnitude but opposite in sign, we can set \( \beta = -\alpha \).

2. **Sum of the Roots**:
According to Vieta's formulas, the sum of the roots is given by:
\[ \alpha + \beta + \gamma = a \]
Substituting \( \beta = -\alpha \):
\[ \alpha - \alpha + \gamma = a \implies \gamma = a \]

3. **Product of the Roots**:
The product of the roots is given by:
\[ \alpha \cdot \beta \cdot \gamma = -c \]
Substituting \( \beta = -\alpha \) and \( \gamma = a \):
\[ \alpha \cdot (-\alpha) \cdot a = -c \implies -\alpha^2 a = -c \implies \alpha^2 a = c \]

4. **Sum of the Products of Roots Taken Two at a Time**:
The sum of the products of the roots taken two at a time is given by:
\[ \alpha \beta + \beta \gamma + \alpha \gamma = b \]
Substituting \( \beta = -\alpha \) and \( \gamma = a \):
\[ \alpha(-\alpha) + (-\alpha)a + \alpha a = b \implies -\alpha^2 - \alpha a + \alpha a = b \implies -\alpha^2 = b \]

5. **Relating the Equations**:
From the previous steps, we have:
- From the product of the roots: \( c = \alpha^2 a \) (Equation 1)
- From the sum of the products of roots: \( b = -\alpha^2 \) (Equation 2)

Substituting Equation 2 into Equation 1:
\[ c = -b a \]
Rearranging gives:
\[ ab = c \]

### Conclusion:
Thus, we conclude that if two roots of the cubic equation are equal in magnitude but opposite in sign, then the relationship between the coefficients is:
\[ ab = c \]