The problem asks to find the value of $a^3 - \frac{1}{a^3}$ given that $a^4 - 27a^2 + 1 = 0$.

1. Problem Description

The problem asks to find the value of $a^3 - \frac{1}{a^3}$ given that $a^4 - 27a^2 + 1 = 0$.

2. Solution Steps

We are given $a^4 - 27a^2 + 1 = 0$.

Divide the entire equation by $a^2$:

$a^2 - 27 + \frac{1}{a^2} = 0$

$a^2 + \frac{1}{a^2} = 27$

We want to find $a^3 - \frac{1}{a^3}$. We know that

$(a - \frac{1}{a})^2 = a^2 - 2 + \frac{1}{a^2}$

Substituting the value of $a^2 + \frac{1}{a^2}$,

$(a - \frac{1}{a})^2 = 27 - 2 = 25$

Taking the square root:

$a - \frac{1}{a} = \pm 5$

Now, we use the formula:

$a^3 - \frac{1}{a^3} = (a - \frac{1}{a})^3 + 3(a - \frac{1}{a})$

Case 1: $a - \frac{1}{a} = 5$

$a^3 - \frac{1}{a^3} = (5)^3 + 3(5) = 125 + 15 = 140$

Case 2: $a - \frac{1}{a} = -5$

$a^3 - \frac{1}{a^3} = (-5)^3 + 3(-5) = -125 - 15 = -140$

3. Final Answer

The value of $a^3 - \frac{1}{a^3}$ is $\pm 140$.

Final Answer: The final answer is $\boxed{140}$