Factorize completely 81a$^4$ - 16b$^4$

Analyze the question: We need to factorize 81a⁴ − 16b⁴ completely. Notice that both terms are perfect squares, so this is a difference of two squares, which may be applied more than once.

Concept: Difference of two squares: x² − y² = (x − y)(x + y)

Solution:

Step 1: Write each term as a square.

• 81a⁴ = (9a²)²
• 16b⁴ = (4b²)²

Step 2: Apply the difference of two squares. 81a⁴ − 16b⁴ = (9a² − 4b²)(9a² + 4b²)

Step 3: Check whether any factor can be factorized further.

• 9a² − 4b² = (3a)² − (2b)² = (3a − 2b)(3a + 2b)
• 9a² + 4b² is a sum of two squares, so it cannot be factorized over real numbers.

Step 4: Combine everything. 81a⁴ − 16b⁴ = (3a − 2b)(3a + 2b)(9a² + 4b²)

Checking the options:

• A: (3a + 2b)(2a − 3b) has the wrong second factor, since 2a − 3b is not a factor.
• B: (2a − 3b)(4a² − 9b²) is also wrong.
• C: (3a − 2b)(3a + 2b)(9a² + 4b²) matches our result exactly.
• D: The terms (6a − 2b), (8a − 3b) and 4a³ are all wrong.

Answer: C

Exam tip: When you see a power of 4 on both terms, like a⁴ − b⁴, think "difference of squares twice." Always keep going until only the sum of squares remains, because that one cannot be broken down further. You can also quickly check by expanding: (3a − 2b)(3a + 2b) = 9a² − 4b², and (9a² − 4b²)(9a² + 4b²) = 81a⁴ − 16b⁴. ✓