Question

Find the factors of $(a+b+c{)}^4-(b+c{)}^4-(c-a{)}^4-(a+b{)}^4+a^4+b^4+c^4$.

Answer 1

Given expression is ${(a+b+c)}^4-{(b+c)}^4-{(c+a)}^4-{(a+b)}^4+a^4+b^4+c^4$

Put $b=0$ in the given expression,
$(a+0+c{)}^4-(0+c{)}^4-(c+a{)}^4-(a+0{)}^4+a^4+0^4+c^4=0$
Similarly, we get expression value as '$0$' for $a=0$ and $c=0$.
Put $a+b+c=0,c+a=-b,b+a=-c,b+c=-a$ in the given expression
$(0{)}^4-(-a{)}^4-(-b{)}^4-(-c{)}^4+a^4+b^4+c^4=0$
As it is fourth power equation, we can it write as
$(a+b+c{)}^4-(b+c{)}^4-(c+a{)}^4-(a+b{)}^4+a^4+b^4+c^4=kabc(a+b+c)$
Put $a=1,b=2,c=3,$ we get
$(1+2+3{)}^4-(2+3{)}^4-(3+1{)}^4-(1+2{)}^4+1^4+2^4+3^4=k\times 1\times 2\times 3\times (1+2+3)$
On simplification, we get
$432=k\left(36\right)$
$k=12$
Therefore, factors are $12abc(a+b+c)$