Question

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

Answer 1

let $f\left(bc\right)=(bc+ca+ab{)}^3-b^3{c}^3-c^3{a}^3-a^3{b}^3$
$f(-ac)=(-ac+ca+ab{)}^3-(-ac{)}^3-c^3{a}^3-a^3{b}^3$
$=a^3{b}^3+a^3{c}^3-a^3{c}^3-a^3{b}^3=0$
If $f\left(x\right)$ has $f(-a)=0,$ then $x+a$ is factor of $f\left(x\right)$
Therefore, we can say $bc+ac$ is factor of $f\left(bc\right)$.
Now, $f(-ab)=(-ab+ac+ab{)}^3-(-ab{)}^3-c^3{a}^3-a^3{b}^3$
$=a^3{c}^3+a^3{b}^3-c^3{a}^3-a^3{b}^3=0$
Therefore , we can say $bc+ab$ is factor of $f\left(bc\right)$.
Similarly, $ab+ac$ is also factor.
Since it is a cubic equation, we can write it as
$(bc+ca+ab{)}^3-b^3{c}^3-c^3{a}^3-a^3{b}^3=k(bc+ac\left)\right(bc+ab\left)\right(ab+ac)$=$kabc(b+c)(c+a)(a+b)$
Put $a=1,b=2,c=3$
$(2\times 3+3\times 1+1\times 2{)}^3-2^3{3}^3-3^3{1}^3-1^3{2}^3=k\times 1\times 2\times 3(2+3\left)\right(1+3\left)\right(1+2)$
On simplification, we get
$108=k\left(36\right)$
$k=\frac{108}{36}=3$
Thus, the factors are $3abc(b+c)(a+c)(a+b)$.