Question

y^3b-64 for factor completely?

Answer 1

use the difference of two cubes:\[y^{3b}-64=(y^b)^3-4^3=(y^b-4)((y^b)^2+(y^b)(4)+4^2)\]\[=(y^b-4)(y^{2b}+4y^b+16)\]

Answer 2

what method are u using?

Answer 3

im using the geometric formula. \[a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\ldots + ab^{n-2}+b^{n-1})\]in this case n = 3

Answer 4

thanks