The number of real roots of  (x−1)4+(x+1)4=16  is

The given equation is (x−1)4+(x+1)4=16 ⇒(x2+1−2x)(x2+1−2x)+(x2+1+2x)(x2+1+2x)=16                                           (or using (a+b)4=a4+4a3b+6a2b2+4ab3+b4) ⇒  2(x4+6x2+1)=16 ⇒  x4+6x2−7=0 ⇒  (x2−1)(x2+7)=0 ⇒  x=±1  are the only real solutions.