If `P + P! = P^3` , then the value of P is :

Options

A. 4
B. 6
C. 0
D. 5

Answer

To solve the equation \( P + P! = P^3 \), we can follow these steps:

### Step 1: Understand the equation
We start with the equation:
\[ P + P! = P^3 \]
Here, \( P! \) (P factorial) is the product of all positive integers up to \( P \).

### Step 2: Rewrite \( P! \)
Recall that \( P! = P \times (P-1)! \). However, for simplicity, we can evaluate specific integer values of \( P \) to find a solution.

### Step 3: Try integer values for \( P \)
We will test integer values for \( P \) from the options provided: 0, 4, 5, and 6.

#### Testing \( P = 0 \):
\[ 0 + 0! = 0^3 \]
Calculating:
\[ 0 + 1 = 0 \quad \text{(since } 0! = 1\text{)} \]
This simplifies to:
\[ 1 \neq 0 \]
So, \( P = 0 \) is not a solution.

#### Testing \( P = 4 \):
\[ 4 + 4! = 4^3 \]
Calculating:
\[ 4 + 24 = 64 \quad \text{(since } 4! = 24\text{)} \]
This simplifies to:
\[ 28 \neq 64 \]
So, \( P = 4 \) is not a solution.

#### Testing \( P = 5 \):
\[ 5 + 5! = 5^3 \]
Calculating:
\[ 5 + 120 = 125 \quad \text{(since } 5! = 120\text{)} \]
This simplifies to:
\[ 125 = 125 \]
So, \( P = 5 \) is a solution.

#### Testing \( P = 6 \):
\[ 6 + 6! = 6^3 \]
Calculating:
\[ 6 + 720 = 216 \quad \text{(since } 6! = 720\text{)} \]
This simplifies to:
\[ 726 \neq 216 \]
So, \( P = 6 \) is not a solution.

### Conclusion
The only value of \( P \) that satisfies the equation \( P + P! = P^3 \) is:
\[ \boxed{5} \]