Factor out:
64x^3 + 1
It is a binomial times a trinomial
(4x+1)(16x^2-4x+1) should work. Check my math again!
[8x^(3/2) - i][8x^(3/2) + i]
Although my previous answer works (if i = sqrt-1), Taasha's is what you are looking for.
To factor the expression 64x^3 + 1, we can use the sum of cubes formula. The sum of cubes formula states that a^3 + b^3 can be factored as (a + b)(a^2 - ab + b^2).
In this case, a is 4x and b is 1, so we have:
(4x)^3 + 1^3 = (4x + 1)(16x^2 - 4x + 1)
So, the correct factored form of 64x^3 + 1 is (4x + 1)(16x^2 - 4x + 1).
Now, let's check your proposed answer: (8x^(3/2) - i)(8x^(3/2) + i).
It seems like you are using the difference of squares formula, which factors a^2 - b^2 as (a + b)(a - b). However, in this case, 64x^3 + 1 cannot be factored using the difference of squares formula because it is not in the form of a difference of squares.
So, while your answer (8x^(3/2) - i)(8x^(3/2) + i) is an interesting factorization, it is not the correct factorization for the given expression. The correct factorization is (4x + 1)(16x^2 - 4x + 1).