The product of the roots of the equation whose roots are greater by unity than the equation `x^(3)-5x^(2)+6x-3=0` is equal to

Options

A. 3
B. 12
C. 15
D. 18

Answer

To solve the problem, we need to find the product of the roots of a new cubic equation whose roots are each greater by one than the roots of the given equation \(x^3 - 5x^2 + 6x - 3 = 0\).

### Step-by-step Solution:

1. **Identify the given equation**:
The given equation is:
\[ x^3 - 5x^2 + 6x - 3 = 0 \]

2. **Let the roots of the given equation be \(r_1, r_2, r_3\)**:
We denote the roots of the given cubic equation as \(r_1, r_2, r_3\).

3. **Define the roots of the new equation**:
According to the problem, the roots of the new equation are \(r_1 + 1, r_2 + 1, r_3 + 1\).

4. **Find the product of the roots of the new equation**:
The product of the roots of the new equation can be expressed as:
\[ (r_1 + 1)(r_2 + 1)(r_3 + 1) \]

5. **Expand the product**:
Using the expansion, we have:
\[ (r_1 + 1)(r_2 + 1)(r_3 + 1) = r_1 r_2 r_3 + (r_1 r_2 + r_2 r_3 + r_3 r_1) + (r_1 + r_2 + r_3) + 1 \]

6. **Use Vieta's formulas**:
From Vieta's formulas for the cubic equation \(x^3 - 5x^2 + 6x - 3 = 0\):
- The sum of the roots \(r_1 + r_2 + r_3 = 5\)
- The sum of the products of the roots taken two at a time \(r_1 r_2 + r_2 r_3 + r_3 r_1 = 6\)
- The product of the roots \(r_1 r_2 r_3 = 3\)

7. **Substitute the values into the expansion**:
Now substituting these values into our expanded product:
\[ (r_1 + 1)(r_2 + 1)(r_3 + 1) = 3 + 6 + 5 + 1 \]

8. **Calculate the final result**:
\[ 3 + 6 + 5 + 1 = 15 \]

Thus, the product of the roots of the new equation is **15**.