Given equation (a2−bc)x2+2(b2−ac)x+(c2−ab)=0 The given roots are equal, then D must be zero. ∴ D = 0 i.e., [2(b2−ac)]2−4(a2−bc)(c2−ab)=0 ⇒ 4(b4+a2c2−2ab2c)−4(a2c2−bc3−a3b+ab2c)=0 ⇒ 4b4+4a2c2−8ab2c−4a2c2+4bc3+4a3b−4ab2c=0 ⇒ 4b4−12ab2c+4bc3+4a3b=0 ⇒ b3+c3+a3−3abc=0 ∴ a3+b3+c3=3ab