Answer:

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Answer:

We know the corollary: if a+b+c=0 then a3+b3+c3=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 p3(q−r)3+q3(r−p)3+r3(p−q)3 can be factorised as follows:

p3(q−r)3+q3(r−p)3+r3(p−q)3=3[p(q−r)×q(r−p)×r(p−q)]=3pqr(q−r)(r−p)(p−q)

Hence, p3(q−r)3+q3(r−p)

Step-by-step explanation:

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