Question

Factorise:
$p^3{(q-r)}^3+q^3{(r-p)}^3+r^3{(p-q)}^3$

Answer 1

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

Using the above corollary taking $a=p(q-r)$, $b=q(r-p)$ and $c=r(p-q)$, we have $a+b+c=p(q-r)+q(r-p)+r(p-q)=pq-pr+qr-pq+pr-qr=0$ then the equation $p^3(q-r{)}^3+q^3(r-p{)}^3+r^3(p-q{)}^3$ can be factorised as follows:

$p^3(q-r{)}^3+q^3(r-p{)}^3+r^3(p-q{)}^3=3[p(q-r)\times q(r-p)\times r(p-q)]=3pqr(q-r\left)\right(r-p\left)\right(p-q)$

Hence, $p^3(q-r{)}^3+q^3(r-p{)}^3+r^3(p-q{)}^3=3pqr(q-r\left)\right(r-p\left)\right(p-q)$