Each of the roots of the equation `x^(3) - 6x^(2) + 6x - 5 = 0` are increased by k so that the new transformed equation does not contain `x^(2)` term. Then k =

Options

A. a) `1/3`
B. b) `1/2`
C. c) `1`
D. d) `-2`

Answer

To solve the problem, we need to find the value of \( k \) such that when each root of the polynomial equation \( x^3 - 6x^2 + 6x - 5 = 0 \) is increased by \( k \), the new equation does not contain the \( x^2 \) term.

### Step-by-step Solution:

1. **Identify the original equation and its roots**:
The original equation is:
\[ x^3 - 6x^2 + 6x - 5 = 0 \]
Let the roots of this equation be \( \alpha, \beta, \gamma \).

2. **Use Vieta's formulas**:
According to Vieta's formulas, for a cubic equation \( ax^3 + bx^2 + cx + d = 0 \):
- The sum of the roots \( \alpha + \beta + \gamma = -\frac{b}{a} \).
Here, \( a = 1 \) and \( b = -6 \):
\[ \alpha + \beta + \gamma = -\frac{-6}{1} = 6 \]

3. **Transform the roots**:
We increase each root by \( k \):
- New roots will be \( \alpha + k, \beta + k, \gamma + k \).

4. **Find the sum of the new roots**:
The sum of the new roots is:
\[ (\alpha + k) + (\beta + k) + (\gamma + k) = \alpha + \beta + \gamma + 3k \]
Substituting the value from Vieta's:
\[ 6 + 3k \]

5. **Set the sum of the new roots to zero**:
Since the new equation should not contain the \( x^2 \) term, the sum of the roots must equal zero:
\[ 6 + 3k = 0 \]

6. **Solve for \( k \)**:
Rearranging the equation gives:
\[ 3k = -6 \]
Dividing both sides by 3:
\[ k = -2 \]

### Final Answer:
The value of \( k \) is:
\[ \boxed{-2} \]