`(16a^(4)-81b^(4))/(8a^(3)+12a^(2)b+18ab^(2)+27b^(3))`
Which of the following expresssion is equivalent to the expression above?

Options

A. `4a^(2)b+9ab^(2)-a^(2)b^(2)`
B. `4a^(2)b-9ab^(2)`
C. `2a+3b`
D. `2a-3b`

Answer

To solve the expression \((16a^{4}-81b^{4})/(8a^{3}+12a^{2}b+18ab^{2}+27b^{3})\), we will factor both the numerator and the denominator separately.

### Step 1: Factor the Numerator
The numerator is \(16a^{4} - 81b^{4}\). We can recognize this as a difference of squares:
\[ 16a^{4} = (4a^{2})^{2} \quad \text{and} \quad 81b^{4} = (9b^{2})^{2} \]
Using the identity \(m^2 - n^2 = (m+n)(m-n)\), we can factor the numerator:
\[ 16a^{4} - 81b^{4} = (4a^{2} + 9b^{2})(4a^{2} - 9b^{2}) \]

### Step 2: Factor Further
Next, we can factor \(4a^{2} - 9b^{2}\) again as it is also a difference of squares:
\[ 4a^{2} - 9b^{2} = (2a)^{2} - (3b)^{2} = (2a + 3b)(2a - 3b) \]
Thus, the complete factorization of the numerator is:
\[ 16a^{4} - 81b^{4} = (4a^{2} + 9b^{2})(2a + 3b)(2a - 3b) \]

### Step 3: Factor the Denominator
Now, we will factor the denominator \(8a^{3} + 12a^{2}b + 18ab^{2} + 27b^{3}\). We can group the terms:
\[ (8a^{3} + 12a^{2}b) + (18ab^{2} + 27b^{3}) \]
Factoring out the common factors from each group:
\[ 4a^{2}(2a + 3b) + 9b^{2}(2a + 3b) \]
Now, we can factor out the common binomial factor \((2a + 3b)\):
\[ = (2a + 3b)(4a^{2} + 9b^{2}) \]

### Step 4: Rewrite the Expression
Now we can rewrite the original expression with the factored forms:
\[ \frac{(4a^{2} + 9b^{2})(2a + 3b)(2a - 3b)}{(2a + 3b)(4a^{2} + 9b^{2})} \]

### Step 5: Cancel Common Factors
We can cancel the common factors \((2a + 3b)\) and \((4a^{2} + 9b^{2})\) from the numerator and denominator:
\[ = 2a - 3b \]

### Final Result
Thus, the expression simplifies to:
\[ 2a - 3b \]

### Conclusion
The equivalent expression is:
\[ \boxed{2a - 3b} \]