Factorise , `a^3-(1)/(a^3) -2a+(2)/(a)`

Explanation

To factorise the expression \( a^3 - \frac{1}{a^3} - 2a + \frac{2}{a} \), we can follow these steps:

### Step 1: Rewrite the expression
We can rewrite the expression by grouping the terms:
\[ a^3 - \frac{1}{a^3} - 2a + \frac{2}{a} = \left(a^3 - \frac{1}{a^3}\right) + \left(-2a + \frac{2}{a}\right) \]

### Step 2: Factor the first part using the difference of cubes
Recall the formula for the difference of cubes:
\[ x^3 - y^3 = (x - y)(x^2 + xy + y^2) \]
Here, let \( x = a \) and \( y = \frac{1}{a} \). Thus,
\[ a^3 - \frac{1}{a^3} = \left(a - \frac{1}{a}\right)\left(a^2 + a \cdot \frac{1}{a} + \left(\frac{1}{a}\right)^2\right) \]
This simplifies to:
\[ a^3 - \frac{1}{a^3} = \left(a - \frac{1}{a}\right)\left(a^2 + 1 + \frac{1}{a^2}\right) \]

### Step 3: Factor the second part
Now, we can factor the second part:
\[ -2a + \frac{2}{a} = -2\left(a - \frac{1}{a}\right) \]

### Step 4: Combine the factors
Now we can combine the two parts:
\[ \left(a - \frac{1}{a}\right)\left(a^2 + 1 + \frac{1}{a^2}\right) - 2\left(a - \frac{1}{a}\right) \]
Factoring out \( \left(a - \frac{1}{a}\right) \):
\[ = \left(a - \frac{1}{a}\right)\left(a^2 + 1 + \frac{1}{a^2} - 2\right) \]

### Step 5: Simplify the expression inside the parentheses
Now simplify the expression inside the parentheses:
\[ a^2 + 1 + \frac{1}{a^2} - 2 = a^2 - 1 + \frac{1}{a^2} \]
This can be rewritten as:
\[ = \left(a^2 - 1\right) + \frac{1}{a^2} = (a - 1)(a + 1) + \frac{1}{a^2} \]

### Final Step: Write the complete factorization
Thus, the complete factorization of the original expression is:
\[ = \left(a - \frac{1}{a}\right)\left(a^2 - 1 + \frac{1}{a^2}\right) \]

### Final Answer
The final factorized form is:
\[ \left(a - \frac{1}{a}\right)\left(a^2 + 1 + \frac{1}{a^2} - 2\right) \]