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 the equation whose roots are greater by unity than the roots of the given polynomial equation \( x^3 - 5x^2 + 6x - 3 = 0 \).

Let the roots of the given equation be \( \alpha, \beta, \gamma \). The roots of the new equation will then be \( \alpha + 1, \beta + 1, \gamma + 1 \).

### Step 1: Calculate the product of the new roots
The product of the new roots can be expressed as:
\[ (\alpha + 1)(\beta + 1)(\gamma + 1) \]

### Step 2: Expand the product
Using the distributive property, we can expand this product:
\[ (\alpha + 1)(\beta + 1)(\gamma + 1) = \alpha\beta\gamma + \alpha\beta + \alpha\gamma + \beta\gamma + \alpha + \beta + \gamma + 1 \]

### Step 3: Use Vieta's formulas
From Vieta's formulas for the polynomial \( x^3 - 5x^2 + 6x - 3 = 0 \), we know:
- The sum of the roots \( \alpha + \beta + \gamma = 5 \) (which is \(-(-5)/1\)).
- The sum of the products of the roots taken two at a time \( \alpha\beta + \beta\gamma + \gamma\alpha = 6 \).
- The product of the roots \( \alpha\beta\gamma = -(-3) = 3 \).

### Step 4: Substitute the values into the expanded product
Now we substitute these values into our expanded product:
\[ (\alpha + 1)(\beta + 1)(\gamma + 1) = \alpha\beta\gamma + (\alpha\beta + \beta\gamma + \gamma\alpha) + (\alpha + \beta + \gamma) + 1 \]
Substituting the values we found:
\[ = 3 + 6 + 5 + 1 \]

### Step 5: Calculate the final result
Now, we can calculate the final result:
\[ 3 + 6 + 5 + 1 = 15 \]

Thus, the product of the roots of the equation whose roots are greater by unity than the roots of the given polynomial is \( \boxed{15} \).