Question

Factorise the following:
${(a-3b)}^3+{(3b-c)}^3+{(c-a)}^3$

Answer 1

We know the corollary: if $a+b+c=0$ then $a^3+b^3+c^3=3abc$

Using the above corollary taking $a=(a-3b)$, $b=(3b-c)$ and $c=(c-a)$, we have $a+b+c=a-3b+3b-c+c-a=0$ then the value of $(a-3b{)}^3+(3b-c{)}^3+(c-a{)}^3$ is:

$(a-3b{)}^3+(3b-c{)}^3+(c-a{)}^3=3[(a-3b)\times (3b-c)\times (c-a)]=3(a-3b\left)\right(3b-c\left)\right(c-a)$

Hence, $(a-3b{)}^3+(3b-c{)}^3+(c-a{)}^3=3(a-3b\left)\right(3b-c\left)\right(c-a)$