Factorize : `p^3(q-r)^3+q^3(r-p)^3+r^3(p-q)^3`

Explanation

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)×q(r−p)×r(p−q)]=3pqr(q−r)(r−p)(p−q)`
Hence, `p^3(q−r)^3+q^3(r−p)^3+r^3(p−q)^3=3pqr(q−r)(r−p)(p−q)`