If p=a+bω+cω2, q=b+cω+aω2, and r=c+aω+bω2, where a,b,c≠0 and ω is the complex cube root of unity, then

p+q+r=a+bω+cω2+b+cω+aω2+c+aω+bω2∴ p+q+r=(a+b+c)1+ω+ω2=0         (1)p, q, r lie on the circle |z| = 2, whose circumcenter is origin. Also, (p + q + r)/3 = 0. Hence the cenroid coincides with circumcenter. So, the triangle is equilateral.Now, (p+q+r)2=0. ⇒ p2+q2+r2=−2pqr1p+1q+1r =−2pqr1a+bω+cω2+1b+cω+aω2+1c+aω+bω2 =−2pqr1ω2aω+bω2+c+1ωbω2+c+aω+1c+aω+bω2 =−2pqraω+bω2+c1ω2+1ω+11=0                  (2)Hence, p2+q2+r2=2(pq+qr+rp)