Factorise :
`16a^(4) + 54a`

Answer

To factorise the expression \( 16a^4 + 54a \), we will follow these steps:

### Step 1: Identify the common factor
First, we need to identify the common factor in the terms \( 16a^4 \) and \( 54a \).

**Common Factor:**
The common factor is \( 2a \).

### Step 2: Factor out the common factor
Now, we will factor out \( 2a \) from the expression:

\[ 16a^4 + 54a = 2a(8a^3 + 27) \]

### Step 3: Recognize the sum of cubes
Next, we notice that \( 8a^3 + 27 \) can be expressed as a sum of cubes. Specifically, we can write:

\[ 8a^3 = (2a)^3 \quad \text{and} \quad 27 = 3^3 \]

So, we have:

\[ 8a^3 + 27 = (2a)^3 + 3^3 \]

### Step 4: Apply the sum of cubes formula
We can use the sum of cubes formula, which states:

\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]

In our case, \( a = 2a \) and \( b = 3 \).

Applying the formula:

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

### Step 5: Simplify the expression
Now, we simplify \( (2a)^2 - (2a)(3) + 3^2 \):

\[ (2a)^2 = 4a^2, \quad (2a)(3) = 6a, \quad 3^2 = 9 \]

Thus, we have:

\[ (2a)^2 - (2a)(3) + 3^2 = 4a^2 - 6a + 9 \]

### Step 6: Combine everything
Putting it all together, we substitute back into our factorization:

\[ 16a^4 + 54a = 2a(2a + 3)(4a^2 - 6a + 9) \]

### Final Answer
The factorized form of \( 16a^4 + 54a \) is:

\[ \boxed{2a(2a + 3)(4a^2 - 6a + 9)} \]

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