The vertices of a triangle are (pq,1/(pq)),(qr,1/(qr)) and  (rq,1/(rq)),where p,q and r are the roots of the equation y3−3y2+6y+1=0. The coordinates of its centroid are

p,q,r are the roots of equation y3−3y2+6y+1=0. So, p+q+r=3,pq+qr+rp=6, and pqr=-1. Now, the centroid of the triangle is (pq+qr+rp3,1pq+1qr+1rp3)i.e., (pq+qr+rp3,p+q+r3pqr)≡(63,3−3)or (2,−1)