Factorise :
`a^(2) + (1)/(a^(2)) + 2 - 5a - (5)/(a)`.

Answer

To factorise the expression \( a^2 + \frac{1}{a^2} + 2 - 5a - \frac{5}{a} \), we will follow these steps:

### Step 1: Rewrite the expression
Start with the given expression:
\[ a^2 + \frac{1}{a^2} + 2 - 5a - \frac{5}{a} \]

### Step 2: Combine terms
Notice that \( a^2 + \frac{1}{a^2} + 2 \) can be rewritten using the identity \( x^2 + 2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 \) where \( x = a \):
\[ a^2 + 2 + \frac{1}{a^2} = \left(a + \frac{1}{a}\right)^2 \]
Thus, we can rewrite the expression as:
\[ \left(a + \frac{1}{a}\right)^2 - 5a - \frac{5}{a} \]

### Step 3: Factor out common terms
Next, we can factor out \( -5 \) from the terms \( -5a - \frac{5}{a} \):
\[ \left(a + \frac{1}{a}\right)^2 - 5\left(a + \frac{1}{a}\right) \]

### Step 4: Let \( x = a + \frac{1}{a} \)
Let \( x = a + \frac{1}{a} \). Then the expression simplifies to:
\[ x^2 - 5x \]

### Step 5: Factor the quadratic expression
Now we can factor the quadratic expression:
\[ x^2 - 5x = x(x - 5) \]

### Step 6: Substitute back for \( x \)
Substituting back \( x = a + \frac{1}{a} \):
\[ (a + \frac{1}{a})(a + \frac{1}{a} - 5) \]

### Final Answer
Thus, the factorised form of the expression is:
\[ \left(a + \frac{1}{a}\right)\left(a + \frac{1}{a} - 5\right) \]

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