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)?
 A: 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}$$

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

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:
$$
p_4 + 4 e_4 = e_1 p_3 - e_2 p_2 + e_3 p_1 = e_1(p_3+e_3) - e_2 p_2
$$
$$
\begin{align}
\iff a^4+b^4+c^4+d^4 + 4 abcd &=  (a+b+c+d)(a^3+b^3+c^3+d^3+abc+abd+acd+bcd) \\ &\quad - (a^2+b^2+c^2+d^2)(ab+ac+ad+bc+bd+cd) 
\end{align}
$$
