SAT Question : Factorise completely 81a⁴ – 16b⁴

A.(3a + 2b) (2a – 3b) (9a² + 4b²)
B.(3a – 2b) (2a – 3b) (4a² – 9b²)
C.(3a – 2b) (3a + 2b) (9a² + 4b²)
D.(3a – 2b) (2a – 3b) (9a² + 4b²)
E.(3a – 2b) (2a – 3b) (9a²- 4b²)

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Solution
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Applying indices, 81a⁴ – 16b⁴ can be rewritten as (9a²)² – (4b²)²

Applying difference of two squares

(9a²)² – (4b²)²
= (9a² – 4b²)(9a² + 4b²)

Applying difference of two squares again:

(9a² – 4b²)(9a² + 4b²) = ((3a)² – (2b)²)(9a² + 4b²)

= (3a – 2b)(3a + 2b)(9a² + 4b²)

The correct option is C

SOURCE: http://samimath.com/q-and-a-factorisation/

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