how do i factor this equation 64a^6n-b^6n ? the "a" is raised to 6n same as with "b"?

'the "a" is raised to 6n same as with "b"'
Use parentheses around the exponents: 64a^(6n) - b^(6n)

64a⁶ⁿ - b⁶ⁿ is the difference of squares: 64a⁶ⁿ - b⁶ⁿ = (8a³ⁿ)² - (b³ⁿ)²
(8a³ⁿ)² - (b³ⁿ)² = (8a³ⁿ + b³ⁿ) (8a³ⁿ - b³ⁿ)

8a³ⁿ + b³ⁿ is the sum of cubes: (2aⁿ)³ - (bⁿ)³
(2aⁿ)³ + (bⁿ)³ = (2aⁿ + bⁿ) ((2aⁿ)² - 2aⁿbⁿ + (bⁿ)²) = (2aⁿ + bⁿ) (4a²ⁿ - 2aⁿbⁿ + b²ⁿ)

8a³ⁿ - b³ⁿ is the difference of cubes: (2aⁿ)³ - (bⁿ)³
(2aⁿ)³ - (bⁿ)³ = (2aⁿ - bⁿ) ((2aⁿ)² + 2aⁿbⁿ + (bⁿ)²) = (2aⁿ - bⁿ) (4a²ⁿ + 2aⁿbⁿ + b²ⁿ)

64a⁶ⁿ - b⁶ⁿ = (2aⁿ + bⁿ) (4a²ⁿ - 2aⁿbⁿ + b²ⁿ) (2aⁿ - bⁿ) (4a²ⁿ + 2aⁿbⁿ + b²ⁿ)

Observe that both terms are perfect cubes. So write it as a difference of cubes:

(4a^(2n))^3 - (b^(2n))^3

So use difference of cubes formula.

Then observe the first factor you find that way is also a difference of squares and so can be factored some more.

How do i factor this equation 64a^6n-b^6n ? the "a" is raised to 6n same as with "b"?

...then you MUST USE PARENTHESES for CLEAR MEANING !!

64a^(6n) - b^(6n) is the difference of two squares...

[8a^(3n) + b^(3n)][8a^(3n) - b^(3n)]

which is the sum and difference of two CUBES...

goes like:
(x³ + y³) = (x + y)(x² - xy + y²)
(x³ - y³) = (x - y)(x² + xy + y²)

Can you complete the factorization now ? Why not ??

64a^6n-b^6n = 2^6 a^(6n) - b^(6n)= ( 2^3 a^(3n) -b^(3n)) ( 2^3 a^(3n) + b^(3n))=
(2 a^n- b^n)( 2^2 a^(2n) + b^(2n) + 2 a^n b^n (a^n b^n)) (2 a^n + b^n) ( 2^2 a^2n + b^2n + 2 2^2a^nb^n( a^n + b^n))
I think this is the answer.

(a^(3n))^2 = a^(6n)
64 a^(6n) - b^(6n) = (8 a^(3n))^2 - (b^(3n) )^2
x^2-y^2 = (x-y)(x+y)

= (8 a ^(3n) - b^(3n) ) ( 8 a^(3n) + b^(3n) )