# Differene of Qintics

1. Jul 2, 2013

### Tenshou

Of course, we all know about the boring ol' Difference of Squares and even the sum and difference of cubes, but what about the sum or difference of quartics, quintics, sextic or better known as the hextic functions,and what about the septics (I know the smell must be unbearable by now, lol), but alas, I kid you not in asking the big questions, Are there difference of these or even sums? It is very hard to check when the power is prime, with the exception of cubic functions those are easily known from grade school.

I have pulled a few together::
$(a-b)(a+b)=(a^{2}-b^{2})$ [Difference of Squares]
$(a-b)(a^{2}+ab+b^{2}) = (a^{3}-b^{2})$ [Difference of Cubes]
$(a-b)(a^{3}+ab^{2}+ba^{2}+b^{3})$ [Difference of Quarts]
$(a^{2}-b^{2})(a^{4}+a^{2}b^{2}+b^{4})$ [Difference of Hextic]

I implore you, Check these, expand them out and if you find something wrong please tell me! Also, if you know about quintics and septics difference and sums and sums and difference for the others functions greater than the cubic I would like to see them :3

Last edited: Jul 2, 2013
2. Jul 2, 2013

### micromass

In general:

$$a^n - b^n = (a-b) (a^{n-1} + a^{n-2}b + a^{n-3}b^2 + ... + a b^{n-2} + b^{n-1})$$

3. Jul 2, 2013

### Tenshou

Oh thank you micromass! I was trying to look for this for the longest time!!

4. Jul 2, 2013

### Mentallic

Depending on what you want it for, I'd suggest if you have quartics, hextics etc. (any even power) then factor them as

$$a^{2k}-b^{2k} = (a^k-b^k)(a^k+b^k)$$

And then continue from there. For example, if we want to factor $a^6-b^6$ then rather than using the formula presented earlier to just factor out a linear a-b factor and get

$$a^6-b^6 = (a-b)(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5)$$

Rather, express it as $(a^3)^2-(b^3)^2$ and take the difference of two squares to yield

$$(a^3-b^3)(a^3+b^3)$$

And now you can apply the formula from earlier to further factor the first factor, but also, if n is not even then

$$a^n+b^n = (a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^2-...-ab^{n-2}+b^{n-1})$$

So then

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

and

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

So finally,

$$a^6-b^6 = (a-b)(a^2+ab+b^2)(a+b)(a^2-ab+b^2)$$

If you compare the factors from each answer, then you can show that

$$(a+b)(a^2+ab+b^2)(a^2-ab+b^2) \equiv a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5$$

5. Jul 2, 2013

### Tenshou

Indeed, I like the formula for plus it is pretty decent lol and that even thing I just discovered my self I was ecstatic when I found it out, I was even more floored when I found out that it was a little harder to find when the power was prime, that was probably due to the fact that you cannot decompose a prime.

Why not write that in compact notations
$a^{n}-b^{n}=(a-b) \sum_{k=0}^{n-1=m} (a^{m-k}b^{k}) \\ a^n+b^n=(a+b) \sum_{k=0}^{n-1=m} (-1)^{k}(a^{m-k}b^{k})$
I mean that is nice and compact even for the sum formula compact is always nice :P Also, What did you mean by what do I want it for.

Last edited: Jul 2, 2013
6. Jul 2, 2013

### Mentallic

What do primes have to do with this?

We couldn't assume that you understood the summation notation, and it's easier to read when expanded anyway. You've made a few mistakes in your sums though.
Firstly, the summation limits of k=0 up to n-1=m doesn't make much sense considering your dummy variable in the sum is k, and introducing m is unnecessary.
I'm also not sure if your limits are correct based on the incorrect upper limit of your sums.
Finally, you need to have $(-1)^k$ instead, because $-1^k = -(1^k)=-1$

$$a^n-b^n = (a-b)\sum_{k=1}^{k=n}a^{n-k}b^{k-1}\hspace{8 mm} n\in Z, n>0$$
$$a^n+b^n = (a+b)\sum_{k=1}^{k=n}(-1)^ka^{n-k}b^{k-1}\hspace{8 mm}m\in Z, m>0, n=2m-1$$

I mean that in some cases, you'd prefer to have the expression factorized as much as possible, while in other cases, you need the larger factor. The expressions are equivalent, it just depends on what the problem is and what you want to achieve by factoring the difference of two quartics for example.

7. Jul 2, 2013

### Tenshou

When I was trying to work them out by hand I mean, it was impossible for me to start and find a pattern

well m was for simplicity, but I guess that I could just start from one and then subtract it from the b term, yours looks much more clearer, thank you. But why did you set k=n ? I don't understand could you explain a little more.

8. Jul 2, 2013

### Mentallic

Ahh I see. Yeah it's not an easy result to just figure out in your head, it's much more likely that you would accidentally discover it though. You may have caught on to the formula if you ever needed to divide a difference of two squares, two cubes, two quartics etc. by its linear factor.

For example,

$$\frac{x^2-1}{x-1}=x+1$$
$$\frac{x^3-1}{x-1}=x^2+x+1$$
$$\frac{x^4-1}{x-1}=x^3+x^2+x+1$$

You'd start to notice the pattern and by induction it's easy to show that it works for any positive integer n. Thus we would get the result

$$x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$$

And following up on that, since we're familiar with differences of two squares, it might give that person the idea of testing this pattern out for values other than 1.

Judging from the summations you posted, I have a feeling that you may have a misunderstanding of how they work?

$$\sum_{i=1}^{i=n}i = \sum_{i=1}^{n}i = 1+2+3+...+(n-1)+n$$

So basically, the expression you want to sum usually has a dummy variable present in it - in this case it's i and previously it was k, and you begin at i=1 and keep adding an integer each time until you reach i=n.

Do you mean why I set k=n at the top of the summation symbol? Because it works for the formula.
We could have equivalently let another dummy variable $k' = k-1$ so $k = k'+1$ and then we'd have

$$\hspace{4 mm}(a-b)\sum_{k=1}^{k=n}a^{n-k}b^{k-1}$$
$$\equiv (a-b)\sum_{k'+1=1}^{k'+1=n}a^{n-(k'+1)}b^{(k'+1)-1}$$
$$\equiv (a-b)\sum_{k'=0}^{k'=n-1}a^{n-k'-1}b^{k'}$$

And replacing k' for k (we can do that because it's just a dummy variable) we get a result without k=n at the top of the summation:

$$\equiv (a-b)\sum_{k=0}^{n-1}a^{n-k-1}b^{k}$$

9. Jul 2, 2013

### Tenshou

Does it really matter though? I mean it is just notation, after all Einstein invented his own notation, although I am not saying I am einstein. I am trying to say that the only liberal part of mathematics is the "abuse" or use of a new notation. I only used it such that I was able to start at k=0 and move up to m where m is just n -1. so I didn't have to end up writing it twice >.< seems kinda lazy, I know but it is a little bit simpler.

10. Jul 2, 2013

### Mentallic

Einstein invented his own notation? I didn't know that!
Or maybe - just taking a stab in the air here - you meant to refer to Euler?

It's not a commonly accepted notation, so if you use it without any explanation and it's not immediately obvious how it works, then it'll create confusion. However, now I understand what you mean. Just replace the m by k and you'll be back to the mainstream summation notation.

11. Jul 2, 2013

### Tenshou

I mean Einstein summation convention? What kind of notation did Euler create? Also, I see, didn't mean to confuse any one, and what do you mean by replace m by k? m isn't a dummy variable... it is just sitting in for n-1, so I didn't have to write it over and over.