⇒(x6+y6)(x2+y2)>(x4+y4)(x4+y4) ⇒x8+x6y2+x2y6+y8>x8+y8+2x4y4 ⇒x2y2(x4+y4)>2x4y4 ⇒(x4+y4)>2x2y2 ⇒(x4+y4)−2x2y2>0 ⇒(x2−y2)2>0 The above equation (1) can also be written as ⇒(y2−x2)2>0 Thus, x is not always greater than y. Statement I is not sufficient. Using equation (1), we get ⇒{(x+y)(x−y)}2>0 ⇒x+y>0 and x−y>0 Thus, Statement II is also not sufficient ∴ Neither Statement − I nor Statement − II is required to answer the question