The correct option is B 3(p-q)(q-r)(r-p)
Let a=p−q,b=q−r and c=r−p
We see that
a+b+c=(p−q)+(q−r)+(r−p)=0
Now, we know that a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)
if (a+b+c)=0 then a3+b3+c3−3abc=0
∴a3+b3+c3=3abc
⇒(p−q)3+(q−r)3+(r−p)3
=3(p−q)(q−r)(r−p)