Factoring Problem: Error in Homework Statement?

[Government$]

Homework Statement

http://alphacapitalist.com/wp-content/uploads/2012/06/factoringproblem.jpg

Homework Equations

/

The Attempt at a Solution

I tried factoring this but it seems to me that author has made an error somewhere. Either solution to this problem is not 1/a-1 or there is error in the problem itself. Why? Well i tried putting a=2 and using author's solution i should get 1 but when i put it in the problem get 1.714. Any thoughts?

 

[HallsofIvy]

I have no clue what this is supposed to be. For one thing it is too small and too dark to read well. Also, there is a symbol between the last two terms inside the parentheses that looks like ":". What does that mean?

 

[Government$]

":" means divided.

Maybe this will help:
(3/a-1 - 3a^2+3a+3/a^2-1 : a^4 - a/a^3 +1) * a - a^2/3

And the solution is 1/a-1

 

[Bread18]

I did it and got $\frac{1}{a-1}$.

First I would get rid of the divide then remember that $a^3 -1 = (a-1)(a^2 + a + 1)$

See how you go from there.

 

[Government$]

Where do you see a^3 - 1?

 

[micromass]

Why don't you start by showing what you tried to solve the problem??

 

[Ray Vickson]

":" means divided.

Maybe this will help:
(3/a-1 - 3a^2+3a+3/a^2-1 : a^4 - a/a^3 +1) * a - a^2/3

And the solution is 1/a-1

That is incorrect: the answer is 1/(a-1), not (1/a)-1, which is what your written expression means.

RGV

 

[Bread18]

Where do you see a^3 - 1?

$a^4 - a = a(a^3 - 1)$

 

[Government$]

Here is my stab at it:

($\frac{3}{a-1} - \frac{3a^2+3a+3}{a^2-1} : \frac{a^4-a}{a^3+1}$) * $\frac{a-a^2}{3}$ = ($\frac{-3a^2}{a^2-1}$ * $\frac{(a+1)(a^2-a+1)}{a(a-1)(a^2+a+1)}$) * $\frac{a-a^2}{3}$ = $\frac{-a(a^2-a+1)(a-a^2)}{(a-1)(a-1)(a^2+a+1)}$ = $\frac{a^2(a^2-a+1)}{(a-1)(a^2+a+1)}$ = $\frac{a^4 - a^3 + a^2}{a^3 - 1}$

 

[Bread18]

Here is my stab at it:

($\frac{3}{a-1} - \frac{3a^2+3a+3}{a^2-1} : \frac{a^4-a}{a^3+1}$) * $\frac{a-a^2}{3}$ = ($\frac{-3a^2}{a^2-1}$ * $\frac{(a+1)(a^2-a+1)}{a(a-1)(a^2+a+1)}$) * $\frac{a-a^2}{3}$ = $\frac{-a(a^2-a+1)(a-a^2)}{(a-1)(a-1)(a^2+a+1)}$ = $\frac{a^2(a^2-a+1)}{(a-1)(a^2+a+1)}$ = $\frac{a^4 - a^3 + a^2}{a^3 - 1}$

What did you do to get to ($\frac{-3a^2}{a^2-1}$ * $\frac{(a+1)(a^2-a+1)}{a(a-1)(a^2+a+1)}$)?

 

[Government$]

OMFG i made such a stuipid mistake. Instead of first dividing i have subtracted and of course i couldn't get a right solution. I have caluclated and gotten a rght solution.
Thank you all!

 

[dimension10]

Homework Statement

http://alphacapitalist.com/wp-content/uploads/2012/06/factoringproblem.jpg

Homework Equations

/

The Attempt at a Solution

I tried factoring this but it seems to me that author has made an error somewhere. Either solution to this problem is not 1/a-1 or there is error in the problem itself. Why? Well i tried putting a=2 and using author's solution i should get 1 but when i put it in the problem get 1.714. Any thoughts?

The symbol ":" has been used in 2 different contexts here. Assuming it means $\divide$, try cancelling out the "a-1".