Factorise
` 81 x ^(4) - 16y^(4)`

Answer

To factorize the expression \( 81x^4 - 16y^4 \), we can follow these steps:

### Step 1: Identify the expression
We start with the expression:
\[ 81x^4 - 16y^4 \]

### Step 2: Recognize it as a difference of squares
Notice that both \( 81x^4 \) and \( 16y^4 \) are perfect squares:
\[ 81x^4 = (9x^2)^2 \quad \text{and} \quad 16y^4 = (4y^2)^2 \]
Thus, we can rewrite the expression as:
\[ (9x^2)^2 - (4y^2)^2 \]

### Step 3: Apply the difference of squares formula
We can use the difference of squares identity:
\[ a^2 - b^2 = (a + b)(a - b) \]
Here, let \( a = 9x^2 \) and \( b = 4y^2 \). Applying the formula gives us:
\[ (9x^2 + 4y^2)(9x^2 - 4y^2) \]

### Step 4: Factor the second term further
Now, we notice that \( 9x^2 - 4y^2 \) is also a difference of squares:
\[ 9x^2 - 4y^2 = (3x)^2 - (2y)^2 \]
Applying the difference of squares formula again:
\[ (3x + 2y)(3x - 2y) \]

### Step 5: Combine all factors
Now we can combine all the factors:
\[ 81x^4 - 16y^4 = (9x^2 + 4y^2)(3x + 2y)(3x - 2y) \]

### Final Answer
Thus, the complete factorization of \( 81x^4 - 16y^4 \) is:
\[ (9x^2 + 4y^2)(3x + 2y)(3x - 2y) \]

---