8x^3 -1
(2x)^3 - 1^3
(2x-1)(2x^2 + 2x + 1^2) -> but this doesnt factor
Since a^3 - b^3 = (a-b)(a^2 + ab + b^2)
Substitute a = 2x and b = 1
8x^3 -1 = (2x - 1)(4x^2 + 2x + 1)
You are right that the second term cannot be factored into monomials with real coefficients.
Thats what I thought but I was not sure. Thanks so much.
To factor the expression 8x^3 - 1, we can use the formula for the difference of cubes:
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
In this case, a is 2x and b is 1. Therefore, we have:
(2x)^3 - 1^3 = (2x - 1)((2x)^2 + (2x)(1) + 1^2)
Now let's simplify this:
(2x)^2 = 4x^2
(2x)(1) = 2x
1^2 = 1
Substituting these values, we get:
(2x - 1)(4x^2 + 2x + 1)
So, the expression 8x^3 - 1 can be factored as (2x - 1)(4x^2 + 2x + 1).
It seems you used 1^2 inside the second term of the factored expression. However, 1^2 is just 1, so the expression should be (2x - 1)(2x^2 + 2x + 1). The factorization (2x - 1)(2x^2 + 2x + 1) is correct and cannot be further simplified.