Find a cubic equation in which each root is greater by unity than a root of the equation `x^3-5x^2+6x-3=0`. Which of the following could be the required equation?

Options

A. `p^3-8p^2+19p-15=0`
B. `p^3-19p^2+8p-15=0`
C. `p^3-8p^2+19p-15=0`
D. `p^3-19-8p+15=0`

Answer

To find a cubic equation in which each root is greater by unity than a root of the given equation \( x^3 - 5x^2 + 6x - 3 = 0 \), we can follow these steps:

### Step 1: Define the original function
Let \( f(x) = x^3 - 5x^2 + 6x - 3 \).

### Step 2: Substitute the new variable
We want to find a new function \( g(y) \) such that the roots of \( g(y) = 0 \) are each 1 greater than the roots of \( f(x) = 0 \). To do this, we substitute \( x = y - 1 \) into \( f(x) \).

### Step 3: Substitute \( x = y - 1 \) into \( f(x) \)
Now, we calculate \( f(y - 1) \):
\[ f(y - 1) = (y - 1)^3 - 5(y - 1)^2 + 6(y - 1) - 3 \]

### Step 4: Expand \( f(y - 1) \)
Now we expand each term:
1. \( (y - 1)^3 = y^3 - 3y^2 + 3y - 1 \)
2. \( -5(y - 1)^2 = -5(y^2 - 2y + 1) = -5y^2 + 10y - 5 \)
3. \( 6(y - 1) = 6y - 6 \)

Combining these, we have:
\[ f(y - 1) = (y^3 - 3y^2 + 3y - 1) + (-5y^2 + 10y - 5) + (6y - 6) - 3 \]

### Step 5: Combine like terms
Now, we combine all the terms:
\[ f(y - 1) = y^3 - 3y^2 - 5y^2 + 3y + 10y + 6y - 1 - 5 - 6 - 3 \]
\[ = y^3 - 8y^2 + 19y - 15 \]

### Step 6: Write the final equation
Thus, we have the cubic equation:
\[ y^3 - 8y^2 + 19y - 15 = 0 \]

### Step 7: Replace \( y \) with \( p \)
To match the standard form, we replace \( y \) with \( p \):
\[ p^3 - 8p^2 + 19p - 15 = 0 \]

### Conclusion
The required cubic equation is:
\[ p^3 - 8p^2 + 19p - 15 = 0 \]