The correct option is B a2+b2c2+d2
We have,
x+iy=√a+ibc+id⇒(x+iy)2=a+ibc+id⇒¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(x+iy)2=¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(a+ibc+id)⇒(x−iy)2=a−ibc−id[∵¯¯¯¯¯¯¯¯¯(z)n=(¯¯¯z)n, ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(z1z2)=¯¯¯¯¯z1¯¯¯¯¯z2]
Therefore,
x4+y4+2x2y2=(x2+y2)2=[(x+iy)(x−iy)]2=(x+iy)2×(x−iy)2=a+ibc+id×a−ibc−id=a2+b2c2+d2