Two ways to do this:
(a) {1-[(d2+d)/2]}2=
{[(2-(d2+d)]/2}2=
[2-(d2+d)]2/4=
[4-4(d2+d)+(d2+d)2]/4=
(4-4d2-4d+d4+2d3+d2)/4=
(d4+2d3-3d2-4d+4)/4.
(b) {1-[(d2+d)/2]}2=1-(d2+d)+[(d2+d)2/4]=
1-d2-d+[(d4+2d3+d2)/4]=
(4-4d2-4d+d4+2d3+d2)/4=(d4+2d3-3d2-4d+4)/4.