# Some formula related with factor of (a+b+c+d)

I am looking for some math formula

For example \begin{align} & a^2 -b^2 = (a+b)(a-b) \\ &a^3 +b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab-bc-ca) \end{align} First one related with factor a+b and the second one related with factor a+b+c then

How about some formula related with a,b,c,d

i.e., is there are some equation factors into (a+b+c+d)?

• Well if $$a^2 - b^2 = (a + b)(a - b)$$ $$\implies (a + b)^2 - (c - d)^2 = (a + b + c + d)(a + b - c - d)$$ That is a formula related with a, b, c, d for example. Is this what you are asking for? $$(a + b)^2 = a^2 + b^2 + 2ab$$ $$\implies (c - d)^2 = c^2 + d^2 - 2cd$$ $$\therefore a^2 + b^2 + c^2 + d^2 + 2(ab - cd) = (a + b + c + d)(a + b - c - d)$$ But you can also do for something like $(a - b)^2 - (c + d)^2$ as well if you want – Mr Pie Sep 12 '17 at 3:36
• @user477343 First term on the RHS should be $a+b+c-d$, shouldn't it? EDIT: but you can modify it to $$(a+b)^2 - (c+d)^2 = (a+b+c+d)(a+b-c-d)$$ – Zubin Mukerjee Sep 12 '17 at 3:37
• Yes sorry about that. But I fixed it up :) EDIT: well technically $x - y - z = x - (y + z)$ so... – Mr Pie Sep 12 '17 at 3:38
• @user477343 $$(a+b) + (c-d) \neq a + b + c + d$$ $$(a+b) - (c-d) \neq a + b + c + d$$ You either want your second term $c-d$ to be instead $c+d$, or $-c-d$, as you have it, I don't think it works – Zubin Mukerjee Sep 12 '17 at 3:47
• Next problem , what is formula for,$$a_1,a_2,a_3\cdots,a_n$$ – neonpokharkar Sep 12 '17 at 4:20

It will be probably $$a^4+b^4+c^4+d^4-b^2a^2-c^2a^2-d^2a^2-b^2c^2-b^2d^2-d^2c^2+16abcd=(a+b+c+d)(a^3+b^3+c^3+d^3-ba^2-ca^2-da^2-ac^2-bc^2-dc^2-ab^2-cb^2-db^2-ad^2-bd^2-cd^2+4bca+4bda+4cad+4bcd)$$

OR

$$\sum a^4 -\sum a^2b^2 +16abcd=(\sum a)(\sum a^3-\sum ab^2 +4\sum abc)$$  $$\sum \text{ represents cyclic summation}$$

• Then is there any other representation for $(a^3+b^3+~$ terms? For example like $a^2+b^2+c^2 - ab-bc-ca = \frac{1}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$ I want to make the rest term positive-definite – phy_math Sep 12 '17 at 4:19
• Let's see, i will try – neonpokharkar Sep 12 '17 at 4:23
• Is it for 3 terms? – neonpokharkar Sep 12 '17 at 4:24

\begin{align} & a^2 -b^2 = (a+b)(a-b) \\ &a^3 +b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab-bc-ca) \end{align}

The two relations are not quite "alike" since the second one is symmetric in $\,a,b,c\,$ (i.e. stays invariant if you permute the variables), while the first one is not (both sides change sign). Maybe a better analog would be for the first relation to be written as $\,a^2+b^2+2ab=(a+b)^2\,$.

With that note, the two equalities duplicate the Newton's identities for $\,n=2\,$ and $\,n=3\,$, respectively (where $p_k$ are the $k^{th}$ power sums, and $e_k$ the elementary symmetric polynomials):

\begin{align} p_2 + 2 e_2 &= e_1 p_1 \\ p_3 - 3 e_3 &= e_1 p_2 - e_2 p_1 = e_1(p_2-e_2) \end{align}

The next identity for $\,n=4\,$ would then be:

$$p_4 + 4 e_4 = e_1 p_3 - e_2 p_2 + e_3 p_1 = e_1(p_3+e_3) - e_2 p_2$$

\begin{align} \iff a^4+b^4+c^4+d^4 + 4 abcd &= (a+b+c+d)(a^3+b^3+c^3+d^3+abc+abd+acd+bcd) \\ &\quad - (a^2+b^2+c^2+d^2)(ab+ac+ad+bc+bd+cd) \end{align}