((4^2x) -63)*((4^x)-64) how do i solve this because of the 63 and 64 not having a common base?

Question

also the same idea with this question? how do i do it?

(2^((2/3)x+1)-3)*(2^((1/3)x) -20)

Answer 1

Add zero to the first term by adding (+1 - 1) - write it as follows:

((4^2x) -63 + 1 - 1)*((4^x)-64)

Add the -1 to the 63:
((4^2x) -64 + 1)*((4^x)-64)

Use the distributive property to split this out as:
[((4^2x) -64)*((4^x)-64)] + [(1)*((4^x)-64)]

This should start you out.

Answer 2

2nd question. Add exponents.

2^(x+1) - 3 (2^(1/3)x -20 (2^(2/3) x + 60

[2^(x + 1) - 3] - 20[2^(2/3)x - 3]

Answer 3

1st question
[(4^2x)-63] *[(4^x)-64]
=(64^x) -(64*16^x) -(63*4^x) +4032