Prove that The sum of two cubes (a3 +b3 ) factors into (a+b)(a2 -ab+b2). Need a little help guys.

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- May 18th 2009, 04:58 PMVonNemo19sum of cubes
Prove that The sum of two cubes (

*a*3 +*b*3 ) factors into (*a*+*b*)(*a*2 -*ab*+*b*2). Need a little help guys. - May 18th 2009, 05:07 PMSodapop
- May 18th 2009, 05:10 PMpickslides

*Can't you just expand $\displaystyle (a + b)(a^2 - ab + b^2)$ to prove this?*

- May 18th 2009, 05:22 PMpickslides
Could be a big task but maybe....

start with $\displaystyle a^3+b^3$ and take out $\displaystyle a+b$ as a factor from both terms.

$\displaystyle (a+b)\frac{a^3}{a+b}+(a+b)\frac{b^3}{a+b}$

$\displaystyle (a+b)\left(\frac{a^3}{a+b}+\frac{b^3}{a+b}\right)$

now divde and simplify the terms inside the taller brackets. - May 18th 2009, 05:29 PMVonNemo19
An just what should I divide each of those terms by, Mr. Soprano?(Wondering)

- May 18th 2009, 05:32 PMVonNemo19
I like the visual, Soda pop. Good lookin' out!

- May 18th 2009, 05:43 PMpickslides
- May 18th 2009, 05:51 PMVonNemo19
I don't quite understand. Doing that would change the value of the expression. And I don't understand your language either. The definition of $\displaystyle \Rightarrow$ is implies, or only if. Isn't that correct. If so, as I said before, I don't understand your implication.

- May 18th 2009, 05:55 PMSodapop
He said long division, you can check this link if you don't remember what it is: http://en.wikipedia.org/wiki/Long_division

- May 18th 2009, 06:41 PMpickslides
Long division will help you (maybe! its all trial and error with proofs) make the equation into a more friendly form.

- May 18th 2009, 06:55 PMVonNemo19
Do me a favor and just do it, man. Please. I tried the polynomial division and I didn't get very far. Anybody just do this. I'm trying to study right now and I know that I coukd eventually figure this out, but I shouuldn't have to. After all, this is MHF!

- May 18th 2009, 07:06 PMSodapop
As I already stated a few weeks ago to someone else, we are here to provide tips, hints and paths for you to solve and answer your own questions. We are not here to do the work for you, we can help you, yes, but you have to learn to do things on your own! Please don't consider this rude from me, as I am only refering to the forum rules.

*10) Do not beg for answers.*

Rest of the rules can be found here: http://www.mathhelpforum.com/math-he...php?do=cfrules

Thanks for your understanding! - May 18th 2009, 07:31 PMVonNemo19
I don't really appreciate someone who has been here less time than I have questioning my motives as to why I need certain answers to given problems. I'll have you know that I taught myself everything that I know about math in prison, and I reached the level of calculus 2. Now, someone who has not experienced what it is like to have to trudge through each page of four different texts with no formal instruction would't understand how hard that is. Now that you know a little bit about me, I'll explain why I need the information I seek.

I'm studying evaluating limits algebraically right now (for the second time, but this time in college) and I ran accross the problem $\displaystyle \lim_{x\to{1}}\frac{x^3-1}{x-1}$. Now I no that the sum of cubes formula id the way to go here, but it occured to me that I don't know how to factor (from scratch) the sum of two cubes. Now this type of stuff bothers me Because I like to know why stuff works, not just how. Now, yes, I could go to wikipedia, do some research, or spend a while with a pencil and paper, but I have something better. MHF! So, if you don't want to help me, or give me the answers to the questions I ask, all you have to do is simply not reply to my thread. The question was stated clearly several times, and I kept recieving ambiguous (to say the least) replies. So, yes, by the fourth time I posted I was growing tired of all of this mystery. But I was not begging.

With all that said, I appreciate your patriotic sentiment towards MHF, and your natural instinct to uphold all of its rules, so for that, iI thank you. - May 18th 2009, 09:32 PMIsomorphism
As it was said before, the easiest way to

an identity is to multiply it out.*verify*

Do you know the distributive law?

$\displaystyle (a^2 - ab + b^2)(a+b) = (a^3 -a^2b + ab^2) + (a^2b - ab^2 + b^3)$

Cancel $\displaystyle ab^2$ and $\displaystyle a^2b$ to get $\displaystyle a^3 + b^3$ - May 18th 2009, 09:45 PMJameson
Now, now children, let's all play nice.

I'm just going to close this thread to avoid any more unneeded confrontation. Remember everyone that helping is a two way street. MHF does not exist to give out answers but those who need help should be given the right amount of info to work it out themselves. Let's all just focus on the common ground that we all want to pursue a higher knowledge of math.