Question

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

Answer 1

Let $f\left(x\right)=(x-a{)}^3(b-c{)}^3+(x-b{)}^3(c-a{)}^3+(x-c{)}^3(a-b{)}^3$
$f\left(a\right)=(a-a{)}^3(b-c{)}^3+(a-b{)}^3(c-a{)}^3+(a-c{)}^3(a-b{)}^3=0$
Similarly, $f\left(b\right)=0$ and $f\left(c\right)=0$.
Therefore, $(x-a)(x-b)(x-c)$ are the factors of the given expression.
Let $f\left(b\right)=f\left(x\right)=(x-a{)}^3(b-c{)}^3+(x-b{)}^3(c-a{)}^3+(x-c{)}^3(a-b{)}^3$
$f\left(c\right)=(x-a{)}^3(c-c{)}^3+(x-c{)}^3(c-a{)}^3+(x-c{)}^3(a-c{)}^3=0$
Therefore, $b-a$ is the factor of $f\left(x\right)$
Similarly we get $(a-b),(c-a)$ as factors of given expression cause of similarity.
since it is a polynomial of sixth degree, we can write as
$(x-a{)}^3(b-c{)}^3+(x-b{)}^3(c-a{)}^3+(x-c{)}^3(a-b{)}^3=k(x-a)(x-b)(x-c)(b-c)(c-a)(a-b)$
Put $a=0,b=1,c=2$
$(x-0{)}^3(1-2{)}^3+(x-1{)}^3(2{)}^3+(x-2{)}^3(-1{)}^3=k(x-0)(x-1)(x-2)(1-0)(2-0)(0-1)$
$-x^3+8(x-1{)}^3-(x-2{)}^3=k\left(x\right)(x-1)(x-2)(-2)$
We know that, if $a+b+c=0,$ then $a^3+b^3+c^3=3abc$
Here, we can write the above expression as,
$(-x{)}^3+(2(x-1){)}^3+(-(x-2){)}^3=3(-x\left)\right(x-1\left)\right(x-2\left)\right(2)$
As, $-x+2(x-1)+(-(x-2\left)\right)=0$
Hence, $k=3$
Therefore, the required factors are $3(x-a)(x-b)(x-c)(b-c)(c-a)(a-b)$.