x^3-8/x^2-4 divided by (x^2+2x+4)/(x^3+8). This becomes ((x-2)(x^2+2x+4)/(x+2)(x-2) times ((x+2)(x^2-2x+4))/(x^2+2x+4). I understand that (x-2)(x^2+2x+4)=(x^3-8) when I distribute it, however, if I didn't have the help of my book, I'd have no clue how to come up with this on my own. Is there a general rule for how to distribute x's to the third power? Thank you!!
yes!
We can factor both the sum and the difference of cubes
A^3 + b^3 = (A+B)(A^2 - AB + B^2) and
A^3 - b^3 = (A-B)(A^2 + AB + B^2)
that is where those factors came from.
Most of the stuff canceled and you should have had x^2 - 2x + 4 as the answer.
Yes, there is a general rule for distributing x's to the third power. It is based on the concept of the distributive property, which states that you multiply each term in one expression by each term in another expression.
Let's break down the example you provided to understand the distribution of x's to the third power.
We have the expression x^3 - 8. This can be written as (x - 2)(x^2 + 2x + 4) by using the method of factoring the difference of cubes. The difference of cubes formula is a^3 - b^3 = (a - b)(a^2 + ab + b^2), where a is x and b is 2.
To understand how this formula is derived, start with the expanded form of the difference of cubes:
(a - b)(a^2 + ab + b^2) = a(a^2 + ab + b^2) - b(a^2 + ab + b^2)
= a^3 + a^2b + ab^2 - a^2b - ab^2 - b^3
= a^3 - b^3
In this case, a = x and b = 2, so we get x^3 - 2^3 = (x - 2)(x^2 + 2x + 4)
So, the general rule for distributing x's to the third power is to use the difference of cubes formula for the expression you have.
Remember, this method is specific to the difference of cubes situation. There are other methods for factoring expressions with different powers of x, such as the difference of squares or perfect square trinomial formulas.