can someone explain to me how i should factor this problem?
x^6-64
How do I know if i'm supposed to use the difference of two squares, or the difference of cubes?
Start with (x^3 - 8)(x^3 + 8).
One factor of x^3 -8 is (x-2). The other is(x^3 -8)/(x-2) = x^2 + 2x + 4.
The other factors are complex numbers, such as x + 2i. I assume they are not wanted.
You are left with
(x-2)(x^2 + 2x + 4)(x^3 + 8)
To factor the expression x^6 - 64, we need to determine whether it should be factored using the difference of two squares or the difference of cubes. Here's how you can figure it out:
1. Difference of two squares: This method is used when you have an expression in the form "a^2 - b^2". In this case, x^6 - 64 does not match the form "a^2 - b^2" because neither x^6 nor 64 are perfect squares.
2. Difference of cubes: This method is used when you have an expression in the form "a^3 - b^3". In this case, x^6 - 64 can actually be factored using the difference of cubes because 64 is a perfect cube.
Now, let's factor x^6 - 64 using the difference of cubes. The difference of cubes formula is:
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
Here, a = x^2 and b = 4. Plugging these values into the formula, we get:
x^6 - 64 = (x^2 - 4)(x^4 + 4x^2 + 16)
So, the factored form of x^6 - 64 is (x^2 - 4)(x^4 + 4x^2 + 16).
In summary, you determined that x^6 - 64 can be factored using the difference of cubes because 64 is a perfect cube. By applying the difference of cubes formula, you found the correct factored form.