Question

[(y^3-1)/(y^2-1)]/ [(y^2+y+1)/(y^2+2y+1)] = ?

Answer 1

\[(y^3-1)/(y^2-1) *( y^2+2y+1)/(y^2+y+1)\]

Answer 2

\[[(y^3-1)(y+1)(y+1)]/(y+1)(y-1)(y^2+2y+1)\]

Answer 3

jeez. I think i messed that up already. my question is can i do anything more with (y^2+y+1)?

Answer 4

\[\frac{(y^3-1)(y^2-1)}{(y^2+y+1)(y^2+2y+1)}=\frac{(y-1)(y^2+y+1)(y-1)(y+1)}{(y^2+y+1)(y+1)(y+1)}\]

Answer 5

I have that on paper, just messed it up typing it in.

Answer 6

now that we have factored things lets cancel!

Answer 7

how did you get (y^2-1) on top?

Answer 8

\[\frac{(y-1) \not{ (y^2+y+1) }(y-1)\not{(y+1)}}{\not{(y^2+y+1)} \not{(y+1)} (y+1)}\]
because you have y^2-1 on top

Answer 9

ok, hold on... I think I entered it incorrectly.

Answer 10

Oh maybe I can't read your problem very well

Answer 11

I do see a division sign now

Answer 12

\[\frac{\frac{y^3-1}{y^2-1}}{\frac{y^2+y+1}{y^2+2y+1}}\]
is this the problem?

Answer 13

I will just put it in terms of multiplication.
\[(y^3-1)(y^2+2y+1)/(y^2-1)(y^2+y+1)\]

Answer 14

yes, that is the problem.

Answer 15

\[\frac{\frac{y^3-1}{y^2-1}}{\frac{y^2+y+1}{y^2+2y+1}} \cdot \frac{\frac{y^2+2y+1}{y^2+y+1}}{\frac{y^2+2y+1}{y^2+y+1}}= \frac{y^3-1}{y^2-1} \cdot \frac{y^2+2y+1}{y^2+y+1}\]
\[=\frac{(y-1)(y^2+y+1)}{(y-1)(y+1)} \cdot \frac{(y+1)(y+1)}{y^2+y+1}\]

Answer 16

\[=\frac{y^2+y+1}{y+1} \cdot \frac{(y+1)^2}{y^2+y+1}=\frac{(y+1)^2}{y+1}=y+1\]

Answer 17

Could you explain (y-1)(y^2+y+1) to me? Is that the "difference of cubes?"

Answer 18

\[y^3-(1)^3\]
yes this is difference of cubes

Answer 19

Recall
\[a^3-b^3=(a-b)(a^2+ab+b^2)\]

Answer 20

Excellent, that was my issue. Thanks!