If `a ^(2) + (1)/( a ^(2)) = 7 and V _(1) and V _(2)` are the minimum and the maximum values of `a ^(3) + (1)/(a ^(3)),`then `V _(2) - V _(1)` is equal to ?

Options

A. `- 36`
B. 36
C. 72
D. `-18`

Answer

To solve the problem, we need to find the difference between the maximum and minimum values of \( a^3 + \frac{1}{a^3} \) given that \( a^2 + \frac{1}{a^2} = 7 \).

### Step-by-Step Solution:

1. **Start with the given equation**:
\[ a^2 + \frac{1}{a^2} = 7 \]

2. **Use the identity for \( a + \frac{1}{a} \)**:
Recall that:
\[ \left(a + \frac{1}{a}\right)^2 = a^2 + 2 + \frac{1}{a^2} \]
Therefore,
\[ a^2 + \frac{1}{a^2} = \left(a + \frac{1}{a}\right)^2 - 2 \]
Substituting the value we have:
\[ 7 = \left(a + \frac{1}{a}\right)^2 - 2 \]
Rearranging gives:
\[ \left(a + \frac{1}{a}\right)^2 = 9 \]

3. **Solve for \( a + \frac{1}{a} \)**:
Taking the square root:
\[ a + \frac{1}{a} = 3 \quad \text{or} \quad a + \frac{1}{a} = -3 \]

4. **Calculate \( a^3 + \frac{1}{a^3} \)**:
We can use the identity:
\[ a^3 + \frac{1}{a^3} = \left(a + \frac{1}{a}\right)^3 - 3\left(a + \frac{1}{a}\right) \]
- For \( a + \frac{1}{a} = 3 \):
\[ a^3 + \frac{1}{a^3} = 3^3 - 3 \cdot 3 = 27 - 9 = 18 \]
- For \( a + \frac{1}{a} = -3 \):
\[ a^3 + \frac{1}{a^3} = (-3)^3 - 3 \cdot (-3) = -27 + 9 = -18 \]

5. **Identify the minimum and maximum values**:
- Maximum value \( V_2 = 18 \)
- Minimum value \( V_1 = -18 \)

6. **Calculate the difference**:
\[ V_2 - V_1 = 18 - (-18) = 18 + 18 = 36 \]

### Final Answer:
Thus, \( V_2 - V_1 = 36 \).