Long Division?

Question

Hey i am trying to do (1-a^3)/(1-a^2). I have completed the square on the bottom and (a-1) is a factor on the top. The answers should be (1+a+a^2)/(1+a) but i keep getting a remainder dividing a-1 into 1-a^3. Any help appreciated thanks.

Answer 1

I don't know where particularly you are having trouble, but I'm guessing that on one of the steps you made a sign error (since you are subtracting negative numbers, it's easy to do this). I've written out the full form of the long division so you can see the process. As you can see, you do indeed obtain a remainder of zero, which is exactly what you would expect given that the factors cancel.

....... -a² ... -a .... -1
a-1 | -a³ .... 0 ..... 0 ...... 1
.....- -a³ .... a²
....... ------------
.............. -a²..... 0
.............- -a²..... a
............. ------------
.......... ........... -a ..... -1
.......... ..........- -a ..... -1
.......... ..........--------------
.......... .......... .......... 0

Answer 2

The difference of cubes for (1 - a^3) = (1 - a)(1 + a + a^2)
The difference of squares for ((1 - a^2) = (1+ a)(1 - a)
Then, (1 - a) cancels out.
rewritten.
(a + 1) | (a^2 + a + 1)

a(a + 1) = a^2 +a
Therefore, the answer should be a + remainder 1 / (a+1).

Answer 3

factoring the top (1-a^3):

(1-a^3) is the difference of cubes, so you use the formula:

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

So, (1-a^3) = (1-a)(1+a+a^2)
---
(1-a^2) is the difference of cubes, so you use the formula:

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

So, (1-a^2) = (1-a)(1+a)
---

Since you have a (1-a) on top and bottom, those cancel and you're left with:

(1+a+a^2)/(1+a)

Answer 4

(1-a^3)/(1-a^2)=a+(1-a)/(1-a^3)