Verified answer

Don`t you mean x^3 - 8? ? ?

If indeed you meant x^3 - 8, then look for all possible factors of -8:

You have 8, -1; -8, 1; 2, -4; -2, 4

Then you check to see which, if any, will give you a zero value.

If you substitute 2 for x, cube it and subtract 8, you get zero.

If this were an equation, x = 2 would be a solution; but you are looking for factors so you have, instead, x - 2.

If you have done long division, you then divide (x - 2) into x^3 - 8 to get the solution that others have given you.

It also might help to memorize the factors of the difference of two cubes.

If your problem was really x^2 - 8, then you can factor this as (x - sqrt8)(x + sqrt8).

The sqrt of 8 may be simplified as 2 times sqrt 2 and you may substitute this form for sqrt 8.

You factor a perfect cube:

x^3 - y^3 into (x - y)(x^2 + xy + y^2)

and

x^3 + y^3 into (x + y)(x^2 - xy + y^2).

The pattern is the cube root of the first and last times the square of the first , the product of the 2 with signs as above, and the square of the last.

x ³ - 8

(x-2)(x ² + 2x + 4)

x³-8 = x³-2³ = (x-2)(x²2x+4)

:>: