SOLUTION: 64x^6 - 1 Im confused.

Question 213846: 64x^6 - 1
Im confused.
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
Remember, if you saw:
(a^2 - b^2)
It's called "difference of squares"
When you see this you should automatically know that the factors are:
(a+b)(a-b)
.
Similarly:
64x^6 - 1
Can be written as:
(8x^3)^2 - (1)^2
Now, you have a "difference of squares" -- leading you to factor it as:
(8x^3 + 1)(8x^3 - 1)
.
Wait, you're still not done. Notice, the above can be rewritten as:
((2x)^3 + 1^3)((2x)^3 - 1^3)
Now you have a "sum of cubes" and a "difference of cubes... these are special factoring cases:
((2x)^3 + 1^3) can be rewritten as:
(2x + 1)(4x^2 - 2x + 1)
and
((2x)^3 - 1^3) can be rewritten as:
(2x - 1)(4x^2 + 2x + 1)
.
This means instead of:
((2x)^3 + 1^3)((2x)^3 - 1^3)
we can rewrite as:
(2x + 1)(4x^2 - 2x + 1)(2x - 1)(4x^2 + 2x + 1)

Reference, for further reading:
http://www.purplemath.com/modules/specfact.htm
http://www.purplemath.com/modules/specfact2.htm

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