I am unsure how to factor x^3-8. Could you please explain?
x^3-8
This is a standard case of the form
(x³-y³)
=(x-y)(x²+xy+y²)
Can you take it from here?
x^3+-448x?
x³-8
=x³-2³
if you put x=x, and y=2 in the above identity, you will have factorized the expression as the product of a binomial and a trinomial.
Certainly! To factor the expression x^3 - 8, we can use a method called factoring the difference of cubes.
The first step is to recognize that x^3 is a perfect cube and 8 is a perfect cube as well.
We then use the formula a^3 - b^3 = (a - b)(a^2 + ab + b^2).
In our case, a is x and b is 2 because 8 is equal to 2^3.
Substituting the values into the formula, we have:
x^3 - 8 = (x - 2)(x^2 + 2x + 4).
So the factored form of x^3 - 8 is (x - 2)(x^2 + 2x + 4).
And that's how you factor x^3 - 8 using the method of factoring the difference of cubes.