A difficult problem on factorization algebra Factorize $(a+b+c)^7 - a^7-b^7-c^7$.
 A: Hint. Note that $$(x^7-y^7)=(x-y)\left(\sum_{k=0}^6 x^ky^{6-k}\right)$$
which implies that $(a+b)$, $(b+c)$ and $(c+a)$ are factors.
A: When $a=-b$, 
$$(a+b+c)^7-a^7-b^7-c^7=c^7-a^7-(-a)^7-c^7=0$$
By Factor Theorem, $a+b$ is a factor of $(a+b+c)^7-a^7-b^7-c^7$.
Similarly, $b+c$ and $c+a$ are also factors of $(a+b+c)^7-a^7-b^7-c^7$.
As $(a+b+c)^7-a^7-b^7-c^7$ is cyclic and symmetric,
$$(a+b+c)^7-a^7-b^7-c^7=(a+b)(b+c)(c+a)[p(a^4+b^4+c^4)+q(a^2b^2+b^2c^2+c^2a^2)+r(a^3b+a^3c+b^3a+b^3c+c^3a+c^3b)+sabc(a+b+c)]$$
for some constants $p$, $q$, $r$ and $s$.
Comparing the coefficients of $a^6b$, we have
$$7=p$$
Comparing the coefficients of $a^5b^2$, we have
$$\binom{7}{2}=p+r$$
Comparing the coefficients of $a^4b^3$, we have
$$\binom{7}{3}=q+r$$
Comparing the coefficients of $a^4b^2c$, we have
$$\binom{7}{4}\binom{3}{2}=p+q+3r+s$$
Therefore, $p=7$, $q=21$, $r=14$ and $s=35$.
$$(a+b+c)^7-a^7-b^7-c^7=7(a+b)(b+c)(c+a)[a^4+b^4+c^4+3(a^2b^2+b^2c^2+c^2a^2)+2(a^3b+a^3c+b^3a+b^3c+c^3a+c^3b)+5abc(a+b+c)]$$
It remains to show that 
$$g(a,b,c)=a^4+b^4+c^4+3(a^2b^2+b^2c^2+c^2a^2)+2(a^3b+a^3c+b^3a+b^3c+c^3a+c^3b)+5abc(a+b+c)$$ 
is irreducible.
If $\alpha a+\beta b+\gamma c$ is a factor of $g(a,b,c)$, then so are $\alpha a+\beta c+\gamma a$, $\alpha b+\beta a+\gamma c$, $\alpha b+\beta c+\gamma a$, $\alpha c+\beta a+\gamma b$ and  $\alpha c+\beta b+\gamma a$.
If $\alpha,\beta,\gamma$ are mutually distinct, then the product of these factors has degree $6$.
So at least two of $\alpha,\beta,\gamma$ are equal, say $\beta=\gamma$. But then
$$[\alpha a+\beta (b+c)][\alpha b+\beta (c+a)][\alpha c+\beta (a+b)]$$
is a factor of degree $3$ of $g(a,b,c)$ and thus there is one more cyclic and symmetric linear factor left. This factor should be $a+b+c$.
However, when $a=-b-c$,
$$(a+b+c)^7-a^7-b^7-c^7=(b+c)^7-b^7-c^7\ne0$$
and hence $a+b+c$ cannot be a factor of $g(a,b,c)$.
Therefore, $g(a,b,c)$ has no linear factors.
$g(a,b,c)$ either has no factors or only has cyclic and symmetric quadratic factors, i.e.
$$g(a,b,c)=[a^2+b^2+c^2+m(ab+bc+ca)][a^2+b^2+c^2+n(ab+bc+ca)]$$
Comparing the coefficients of $a^3b$,
$$2=m+n$$
Comparing the coefficients of $a^2bc$,
$$5=m+n$$
Such pair of $m$ and $n$ does not exist.
$g(a,b,c)$ is irreducible.
