SOLUTION: I need help solving this problem Factor completely 27b^4-64b I came up with b(-27+8b^3) which is incorrect.... Not sure how to do it.

Question 195803: I need help solving this problem
Factor completely
27b^4-64b
I came up with b(-27+8b^3) which is incorrect.... Not sure how to do it.
Found 2 solutions by nerdybill, Earlsdon:
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
27b^4-64b
Start by factoring out 3b:
b(27b^3 - 64)
Rewrite as:
b((3b)^3 - 4^3)
Notice, you now have a "difference of cubes". This is a "special" factor.
An expression of the form a3 - b3. The difference of two cubes factors into
(a - b)(a^2 + ab + b^2).
.
So we can rewrite:
b((3b)^3 - 4^3)
as
b(3b - 4)((3b)^2 + (3b)(4) + 4^2)
b(3b - 4)(9b^2 + 12b + 16)

Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Factor:
First, factor a (b) to get:
Notice that the parentheses contain the difference of cubes.

The difference of cubes factors thus: , so...

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