1. 1. 8X^3 - 27 = 0

Notice that 8, x^3, and 27 are perfect cube terms. We can use the difference of cubes to solve this:

(a^3 - b^3) = (a - b)(a^2 + ab + b^2)

Notice that a is the cubed root of a^3 and b is the cubed root of b^3, so to solve this problem, we need the cubed roots of the two terms.

8x^3 --> 2x

27 --> 3

Therefore,

8x^3 - 27 = 0

(2x - 3)[(2x)^2 + (2x)(3) + (3)^2] = 0

(2x - 3)(4x^2 + 6x + 9) = 0

2. X^3 - 3X^2 = 16x - 48

First, we need to 'set this equal to 0' by moving everything to the left.

x^3 - 3x^2 - 16x + 48 = 0

Now, I'm going to try to factor this by grouping:

(x^3 - 3x^2) + (-16x + 48) = 0

x^2(x - 3) - 16(x - 3) = 0 ... Notice that (x - 3) is common in both terms, I can factor it to get:

(x - 3)(x^2 - 16) = 0 ... Notice that x^2 - 16 is a difference of squares:

(x - 3)(x + 4)(x - 4) = 0

3. X^4 - 2X - 8 = 0

Check to make sure you didn't write this one wrong. I doubt it should have an x^4 term.

I have to go... hopefully someone else can help solve the rest.