how do I factor out 64-8x^3?

a³-b³ = (a-b)(a²+ab+b²)

64-8x³
= 8(8-x³)
= 8(2³-x³)
= 8(2-x)(2²+2x+x²)
= 8(2-x)(4+2x+x²)

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

64 - 8x^3
= 8(8 - x^3)
= 8(2^3 - x^3)
= 8(2 - x)(2^2 + 2x + x^2)
= 8(2 - x)(4 + 2x + x^2)

both terms are divisible by "8", so you can pull it out of both. that would leave you with

8(8 - x^3)

you can factor further because of the difference of two cubes, leaving the final answer of

8(2 - x)(x^2 + 2x + 4)

(4-2x)(16+8x+4x^2)=
=64+32x+16x^2 -32x-16x^2-8x^3=
=64-8x^3
=2(2-x)(4)(4+2x+x^2)=
=8(2-x)(x^2+2x+4)

rewrite it as 4^3 - (2x)^3 = (4-2x)(16 + 8x + 4x^2)
note: a^3 - b^3 = (a-b)(a^2 + ab + b^2)

64 = 4^3
8 = 2^3

64 - 8x^3 = 4^3 - 2^3 x^3 = 4^3 - (2x)^3

Now, use the difference of cubes identity:
a^3 - b^3 = (a - b)(a^2 + ab + b^2)

a=4, b=2x

64 - 8x^3 = (4 - 2x)(16 + 8x + 4x^2)