The cube of the number `rho` is 16 times the number. Than find `rho` where `rho != 0` and `rho != -4`.

Answer

To solve the problem, we need to find the value of the number \( \rho \) such that the cube of \( \rho \) is 16 times the number itself. We can follow these steps:

### Step 1: Set up the equation
The problem states that the cube of the number \( \rho \) is equal to 16 times the number. We can express this mathematically as:
\[ \rho^3 = 16\rho \]

### Step 2: Rearrange the equation
To solve for \( \rho \), we can rearrange the equation:
\[ \rho^3 - 16\rho = 0 \]

### Step 3: Factor the equation
Next, we can factor out \( \rho \) from the equation:
\[ \rho(\rho^2 - 16) = 0 \]

### Step 4: Further factor the quadratic
The expression \( \rho^2 - 16 \) is a difference of squares, which can be factored further:
\[ \rho(\rho - 4)(\rho + 4) = 0 \]

### Step 5: Solve for \( \rho \)
Now we can set each factor equal to zero:
1. \( \rho = 0 \)
2. \( \rho - 4 = 0 \) → \( \rho = 4 \)
3. \( \rho + 4 = 0 \) → \( \rho = -4 \)

### Step 6: Exclude the invalid solutions
The problem states that \( \rho \neq 0 \) and \( \rho \neq -4 \). Therefore, we can exclude \( \rho = 0 \) and \( \rho = -4 \).

### Conclusion
The only valid solution left is:
\[ \rho = 4 \]