Can someone check my answer?
5x^3 - 40
=5(x^3-8)
=5(x-2)(x^2+2x+4)
Also, how do you factor x^6-y^6 as a difference of cubes?
As a difference of squares, is it (x^3+y^3)(x^3-y^6)?
Yes.
typo: (x^3-y^3)
To check your answer, we can multiply out the factored expression and see if it simplifies back to the original expression.
Let's multiply the factored expression: 5(x-2)(x^2+2x+4).
First, distribute the 5 to both terms inside the parentheses:
5(x-2) = 5x - 10.
Now distribute the remaining factor (x^2+2x+4):
(5x - 10)(x^2+ 2x + 4) = 5x(x^2+ 2x + 4) - 10(x^2+ 2x + 4).
= 5x^3 + 10x^2 + 20x - 10x^2 - 20x - 40.
Combine like terms:
= 5x^3 - 40.
As you can see, the factored expression 5(x-2)(x^2+2x+4) simplifies back to the original expression 5x^3 - 40. Therefore, your answer is correct.
Regarding factoring x^6 - y^6, it cannot be factored as a difference of cubes since neither x^6 nor y^6 can be expressed as perfect cubes.
However, you are correct in recognizing that it can be factored as a difference of squares. The correct factored form is:
x^6 - y^6 = (x^3 + y^3)(x^3 - y^3).
Note that (x^3 + y^3) is the sum of cubes, and (x^3 - y^3) is the difference of cubes.