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.

Algebra ->  Equations -> 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.      Log On


   

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) About Me  (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) About Me  (Show Source):
You can put this solution on YOUR website!
Factor:
27b%5E4-64b First, factor a (b) to get:
b%2827b%5E3-64%29 Notice that the parentheses contain the difference of cubes.
b%28%283b%29%5E3-%284%29%5E3%29
The difference of cubes factors thus: A%5E3-B%5E3+=+%28A-B%29%28A%5E2%2BAB%2BB%5E2%29, so...
b%28%283b%29%5E3-%284%29%5E3%29+=+highlight%28b%283b-4%29%289b%5E2%2B12b%2B16%29%29