The cube of a number x is nine times of x then find x x `ne` 0 and x `ne` -3

Options

A. 8
B. 2
C. 4
D. 3

Explanation

To solve the problem, we need to find the value of \( x \) such that the cube of \( x \) is equal to nine times \( x \), with the conditions that \( x \neq 0 \) and \( x \neq -3 \).

### Step-by-Step Solution:

1. **Set up the equation**:
We start with the statement "the cube of a number \( x \) is nine times \( x \)". This can be written mathematically as:
\[ x^3 = 9x \]

2. **Rearrange the equation**:
To solve for \( x \), we need to rearrange the equation:
\[ x^3 - 9x = 0 \]

3. **Factor out the common term**:
We can factor out \( x \) from the left side:
\[ x(x^2 - 9) = 0 \]

4. **Set each factor to zero**:
This gives us two factors to consider:
- \( x = 0 \)
- \( x^2 - 9 = 0 \)

5. **Solve the second factor**:
We can solve \( x^2 - 9 = 0 \) by recognizing it as a difference of squares:
\[ x^2 - 9 = (x - 3)(x + 3) = 0 \]
This gives us two solutions:
- \( x - 3 = 0 \) → \( x = 3 \)
- \( x + 3 = 0 \) → \( x = -3 \)

6. **Consider the restrictions**:
According to the problem, \( x \) cannot be \( 0 \) or \( -3 \). Therefore, we discard \( x = 0 \) and \( x = -3 \).

7. **Final solution**:
The only valid solution left is:
\[ x = 3 \]

### Summary:
The value of \( x \) that satisfies the equation \( x^3 = 9x \) under the given conditions is:
\[ \boxed{3} \]