ok, so far
But now you are dealing with the sum/difference of cubes
There are two versions
A^3 + B^3 = (A+B)(A^2 - AB + B^2)
and
A^3 - B^3 = (A-B)(A^2 + AB + B^2)
so for the first factor of
b^3 + 8
= (b+2)(b^2 - 2b + 4)
You try the second one.
Your final answer will have 4 factors.
if i factor b^6-64 will I get
(b^3+8)(b^3-8)??
3 answers
SO WILL THE SECOND ONE BE
(b+2)(b^2+2b-4)????
(b+2)(b^2+2b-4)????
close, look at the pattern for
A^3 - B^3 = (A-B)(A^2 + AB + B^2)
vs
(b^3-8) = (b-2)(b^2 + 2b + 4)
A^3 - B^3 = (A-B)(A^2 + AB + B^2)
vs
(b^3-8) = (b-2)(b^2 + 2b + 4)