Factorise :
`2a^(7) - 128a`

Answer

To factorise the expression \( 2a^7 - 128a \), we can follow these steps:

### Step 1: Identify the common factor
First, we observe that both terms in the expression \( 2a^7 \) and \( -128a \) have a common factor. The common factor is \( 2a \).

### Step 2: Factor out the common factor
We factor out \( 2a \) from the expression:

\[ 2a^7 - 128a = 2a(a^6 - 64) \]

### Step 3: Recognize the difference of squares
Next, we notice that \( 64 \) can be expressed as \( 8^2 \) (since \( 64 = 8^2 \)). Thus, we can rewrite \( a^6 - 64 \) as a difference of squares:

\[ a^6 - 64 = a^6 - 8^2 \]

### Step 4: Apply the difference of squares formula
Using the difference of squares formula \( x^2 - y^2 = (x - y)(x + y) \), we can factor \( a^6 - 8^2 \):

\[ a^6 - 8^2 = (a^3 - 8)(a^3 + 8) \]

### Step 5: Factor further using the sum and difference of cubes
Now we can factor \( a^3 - 8 \) and \( a^3 + 8 \) using the sum and difference of cubes formulas:

1. **For \( a^3 - 8 \)** (which is \( a^3 - 2^3 \)):
\[ a^3 - 2^3 = (a - 2)(a^2 + 2a + 4) \]

2. **For \( a^3 + 8 \)** (which is \( a^3 + 2^3 \)):
\[ a^3 + 2^3 = (a + 2)(a^2 - 2a + 4) \]

### Step 6: Combine all factors
Putting it all together, we have:

\[ 2a(a^6 - 64) = 2a((a - 2)(a^2 + 2a + 4)((a + 2)(a^2 - 2a + 4))) \]

Thus, the complete factorization of \( 2a^7 - 128a \) is:

\[ 2a(a - 2)(a + 2)(a^2 + 2a + 4)(a^2 - 2a + 4) \]

### Final Answer:
\[ 2a(a - 2)(a + 2)(a^2 + 2a + 4)(a^2 - 2a + 4) \]

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