if i factor b^6-64 will I get
(b^3+8)(b^3-8)??
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.
SO WILL THE SECOND ONE BE
(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)
Yes, you are correct! If you factor the expression b^6 - 64, you will indeed get (b^3 + 8)(b^3 - 8). Let me explain how you can arrive at that factorization.
To factor b^6 - 64, you can use the difference of squares property. The difference of squares states that a^2 - b^2 can be factored as (a + b)(a - b). In this case, a is b^3 and b is 8.
So, applying the difference of squares, we get:
b^6 - 64 = (b^3)^2 - 8^2
Now, we can recognize that (b^3)^2 is like a^2 and 8^2 is like b^2. We can rewrite the expression as:
(b^3 + 8)(b^3 - 8)
This is the factored form of the original expression b^6 - 64.