Question

Find the factors of ${(a+b+c)}^3-{(b+c-a)}^3-{(c+a-b)}^3-{(a+b-c)}^3$.

Answer 1

Put a=0 in the expression, we get
$(b+c{)}^3-(b+c{)}^3-(c-b{)}^3-(b-c{)}^3=0$
That means $a$ is factor of this expression.
Put $b=0$ in the expression, we get
$(a+c{)}^3-(c-a{)}^3-(a+c{)}^3-(a-c{)}^3$
This shows that $b$ is also a factor of the given expression
Similarly, $c$ is also a factor of the given expression.
Since it is a cubic equation and we already have three roots i.e $a,b,c$
We can write it as $(a+b+c{)}^3-(b+c-a{)}^3-(c+a-b{)}^3-(a+b-c{)}^3=kabc$, let k is some constant.
Put $a=1,b=2,c=3$ on both sides, we get
$(1+2+3{)}^3-(2+3-1{)}^3-(3+1-2{)}^3-(1+2-3{)}^3=k\times 1\times 2\times 3.$
$6^3-4^3-2^3-0^3=k\times 6$
$216-64-8=k\times 6$
$144=k\times 6$
$k=24$
Therefore the factors are $24abc$.