Factorise by grouping method :
`a^(4) - 2a^(3) - 4a + 8`

Answer

To factorise the expression \( a^4 - 2a^3 - 4a + 8 \) by grouping, we can follow these steps:

### Step 1: Group the terms
We can group the terms in pairs:
\[ (a^4 - 2a^3) + (-4a + 8) \]

### Step 2: Factor out the common factors from each group
From the first group \( a^4 - 2a^3 \), we can factor out \( a^3 \):
\[ a^3(a - 2) \]
From the second group \( -4a + 8 \), we can factor out \( -4 \):
\[ -4(a - 2) \]

### Step 3: Write the expression with the factored groups
Now we can rewrite the expression using the factored forms:
\[ a^3(a - 2) - 4(a - 2) \]

### Step 4: Factor out the common binomial factor
Now we see that \( (a - 2) \) is a common factor:
\[ (a - 2)(a^3 - 4) \]

### Step 5: Recognize the difference of cubes
The term \( a^3 - 4 \) can be recognized as a difference of cubes since \( 4 = 2^3 \):
\[ a^3 - 2^3 \]
We can factor this using the difference of cubes formula \( x^3 - y^3 = (x - y)(x^2 + xy + y^2) \):
\[ a^3 - 2^3 = (a - 2)(a^2 + 2a + 4) \]

### Final Factored Form
Putting it all together, we have:
\[ (a - 2)(a - 2)(a^2 + 2a + 4) = (a - 2)^2(a^2 + 2a + 4) \]

Thus, the final factorised form of the expression \( a^4 - 2a^3 - 4a + 8 \) is:
\[ (a - 2)^2(a^2 + 2a + 4) \]

---