1. ## Factoring

Well we had this question on our test today, question number 1 at that, and I just want to know how it's done...

btw, I don't think I did that bad on the test, it's just number 1 was hard. I think.

$27x^3 - 37y^3 - 64$

If someone could show me the solution and its steps that would be superly awsome.

Edit: oops. the question asked to factor this fully.

2. Originally Posted by Freaky-Person
$27x^3 - 37y^3 - 64$
$37y^3=64y^3-27y^3$

I think that's enough to kill the problem.

3. Originally Posted by Krizalid
$37y^3=64y^3-27y^3$

I think that's enough to kill the problem.
why are they all y cubed? O.o

What happened?

Yeah you killed the problem alright, and a piece of my respect really...

4. Are you sure that's the problem?

I think Krizalid did that because he noticed 64-37=27 and then the answer would be $y \in \mathbb{R}$ because $27y^{3}=27y^{3}$

5. Originally Posted by SnipedYou
Are you sure that's the problem?

I think Krizalid did that because he noticed 64-37=27 and then the answer would be $y \in \mathbb{R}$ because $27y^{3}=27y^{3}$
That's the problem EXACTLY. We're supposed to FACTOR it.

$27x^3 - 37y^3 - 64$

6. Well you can get the $y^{3}$ onto its own side and then factor
$37y^{3}=(3x-4)(9x^{2}+12x+16)$ I don't know how to take this any further.

7. Originally Posted by SnipedYou
Well you can get the $y^{3}$ onto its own side and then factor
$37y^{3}=(3x-4)(9x^{2}+12x+16)$ I don't know how to take this any further.
you cannot introduce an equal sign if it was not there in the first place. in doing so you are assuming the formula was equal to zero. we were not given such information

8. Originally Posted by SnipedYou
Well you can get the $y^{3}$ onto its own side and then factor
$37y^{3}=(3x-4)(9x^{2}+12x+16)$ I don't know how to take this any further.
That's so awsome, because that's as far as I got on the test,

question is if it can actually go further. (without the equals, I left it at minus)

9. Originally Posted by Freaky-Person
That's so awsome, because that's as far as I got on the test,

question is if it can actually go further. (without the equals, I left it at minus)
(remember, the equal sign is not supposed to be there).

maybe it's because i'm tired, but i can't see a way to "fully" factor this. it seems you will end up with a plus or a minus sign between terms

10. Originally Posted by Freaky-Person

$27x^3 - 37y^3 - 64$
I constructed the most general possible factor and could not reproduce it. This cannot be factored over the rationals.

-Dan

11. Originally Posted by topsquark
I constructed the most general possible factor and could not reproduce it. This cannot be factored over the rationals.

-Dan
ok, so i'm not crazy

when you get the paper back, make sure you have the correct question, Freaky-Person

12. Originally Posted by Jhevon
ok, so i'm not crazy

when you get the paper back, make sure you have the correct question, Freaky-Person
I swear, that's the question EXACTLY!

I stared at it for like 5 minutes before I realised my teacher is probably evil enough to put something like that on. I stared at the '37' for god knows how long wishing it to be a perfect cube T.T

Thanks for all the help though guys. I guess I'll wait for the results.

13. Originally Posted by Freaky-Person
I swear, that's the question EXACTLY!

I stared at it for like 5 minutes before I realised my teacher is probably evil enough to put something like that on. I stared at the '37' for god knows how long wishing it to be a perfect cube T.T

Thanks for all the help though guys. I guess I'll wait for the results.
yes, and when the result comes, tell your professor, "topsquark says this can't be factored, and he's smarter than you!"...don't call my name though, i have enough enemies

14. Originally Posted by Jhevon
yes, and when the result comes, tell your professor, "topsquark says this can't be factored, and he's smarter than you!"...don't call my name though, i have enough enemies
Ummmm... Thanks Jhevon.

-Dan