If `a^(2) + (1)/(a^(2)) = 18 and a ne 0`, find:
`a^(3)- (1)/(a^(3))`

Answer

To solve the problem where \( a^2 + \frac{1}{a^2} = 18 \) and we need to find \( a^3 - \frac{1}{a^3} \), we can follow these steps:

### Step 1: Find \( a - \frac{1}{a} \)

We know that:
\[ a^2 + \frac{1}{a^2} = (a - \frac{1}{a})^2 + 2 \]
So we can rearrange this to find \( a - \frac{1}{a} \):
\[ (a - \frac{1}{a})^2 = a^2 + \frac{1}{a^2} - 2 \]
Substituting the known value:
\[ (a - \frac{1}{a})^2 = 18 - 2 = 16 \]
Taking the square root:
\[ a - \frac{1}{a} = \sqrt{16} = 4 \quad \text{(since } a \neq 0\text{)} \]

### Step 2: Find \( a^3 - \frac{1}{a^3} \)

We can use the identity:
\[ a^3 - \frac{1}{a^3} = (a - \frac{1}{a}) \left( (a - \frac{1}{a})^2 + 3 \right) \]
Substituting the value we found:
\[ a^3 - \frac{1}{a^3} = 4 \left( 4^2 + 3 \right) \]
Calculating \( 4^2 + 3 \):
\[ 4^2 = 16 \quad \Rightarrow \quad 16 + 3 = 19 \]
Now substituting back:
\[ a^3 - \frac{1}{a^3} = 4 \times 19 = 76 \]

### Final Answer:
Thus, the value of \( a^3 - \frac{1}{a^3} \) is \( \boxed{76} \).

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