(x^3+8)/(x^4-16).

thanks.

The denominator is cleary the differerence of two squares (X^2-4)(x^2+4). That can again be factored to (x+2)(x-2)(x^2+4).

The numerator can be factored using the trinomial simplification..

http://www.algebra-online.com/sum-difference-cubes-1.htm

so (x^2-2x+4)/((x-2)(x^2+4)) is as far as I can go?

yes.

not true

9 months ago

## I apologize for the mistake in my previous response. Let's simplify the expression correctly.

To simplify (x^3 + 8)/(x^4 - 16), we can first factor the numerator and the denominator.

Numerator (x^3 + 8):
The numerator can be factored using the sum of cubes formula, which states that a^3 + b^3 = (a + b)(a^2 - ab + b^2). In this case, a = x and b = 2.

So, x^3 + 8 can be written as (x + 2)(x^2 - 2x + 4).

Denominator (x^4 - 16):
The denominator is actually the difference of squares. It can be written as (x^2)^2 - 4^2, which is (x^2 - 4)(x^2 + 4).

Now, the expression becomes [(x + 2)(x^2 - 2x + 4)] / [(x^2 - 4)(x^2 + 4)].

Next, we can cancel out the common factors in the numerator and denominator. Both the numerator and denominator have (x^2 - 4) as a factor.

Finally, after canceling out (x^2 - 4), we get the simplified expression:

(x + 2) / (x^2 + 4).

Therefore, the simplified expression is (x + 2) / (x^2 + 4).

I apologize again for the earlier mistake.