Need help with Alg. 1 problem...can you help?

Question

http://i134.photobucket.com/albums/q103/thepresident022/mh.jpg

There's a link if you wanna' see it just how it's in the book, or

x^3-1
-------------------
x^3+x^2+x


Ok so far I've hypothesized that maybe you factor an x out and so it would look something like
x(x-1)(x+1)
---------------------
x(x^2+x+1)

But maybe that's wrong? I can smell the answer, but I just can't seem to get it. Any help?

I wish I knew as much math as you do :-)

Answer 1

Simplify: (x^3-1)/(x^3+x^2+x)

First: simplify the numerator - its the difference of cubes ---
The Difference of Cubes Formula is: (a-b)(a^2+ab+b^2)

*Express "x^3 - 1" in lowest terms.....

(x)(x)(x) - (1)(1)(1)

*There are two variables:
a = x
b = 1

*Replace the terms with the corresponding variables in the formula....

(x-1)(x^2+x(1)+(1)^2)
(x-1)(x^2+x+1)

So far, you have....

[(x-1)(x^2+x+1)]/(x^3+x^2+x)

Sec: factor the numerator - find the least common factor which is the "x" variable....

x(x^2+x+1)

Now, you have....

[(x-1)(x^2+x+1)]/[x(x^2+x+1)]

Third: cross cancel "like" terms & your left with....

(x-1)/x

Answer 2

1) (x^3 - 1) / (x^3 + x^2 + x)

Your first step is to factor top and bottom. Note that the top factors as a difference of cubes; a difference of cubes factors in the following fashion: a^3 - b^3 = (a - b) (a^2 + ab + b^2)
x can be factored from the bottom. Therefore, we obtain,

[(x - 1)(x^2 + x + 1)] / [x(x^2 + x + 1)]

Notice the common factor on the top and bottom; we can cancel these, to obtain,

(x - 1) / x

You were correct when factoring the bottom, but when factoring the top, you have to deal with a difference of cubes.

To aid you in how to factor sums and differences of cubes, all you have to do is note the following steps:

1) Take the cube root of each term, and place in the first set of brackets. For x^3 - 1, we have

(x - 1) (? + ? + ?)

You're going to perform these steps:
a) Square the first, (b) Negative product, (c) Square the last.

"Square the first" - square the first value in the first set of brackets. In this case, we're squaring x, so we get x^2.

(x - 1)(x^2 + ? + ?)

"Negative product" - multiply the two terms in the first set of brackets together, and then take their negative. x times (-1) is equal to -x, and the negative of -x is +x, so we get

(x - 1)(x^2 + x + ?)

"Square the last" - square the last term in the first brackets. (-1) squared is 1.

(x - 1)(x^2 + x + 1)

Answer 3

(x-1)(x^2+x+1)
------------------- = (x-1)\ x
x(x^2+x+1)

Answer 4

if you factor x^3-1 i think its = x(x^2-1) and x^3+x^2+x is = x(x^2+x+1)..so i think it will look like this x(x^2-1)
------------ x(x^2+x+1) then elminate the x.. then (x^2)(x^2+x+1) -1(x^2+x+1)=x^4+x^3+x^2-x^2+x+1
=x^4+x^3+x+1...if this is wrong anybody..just correctin it pls.

Answer 5

WRITE X^3-1 AS (X-1)(X^2+X+1).CANCEL THE LARGER TERM. YOU'RE LEFT WITH (X-1)/X.

Answer 6

x^3-1=(x-1)(x^2+x+1) I think formula is not wrong
then result is
1-1/x :)