-(c+d)^3+3ab(c+d)+(c+d)^3-3cd(c+d)-3(a+d)(b+d)(c+d)
=【-(c+d)^3+(c+d)^3】+3ab(c+d)-3cd(c+d)-3(a+d)(b+d)(c+d)
=3{ab(c+d)-cd(c+d)-(a+d)(b+d)(c+d)}
=3{(c+d)ab-(c+d)cd-(c+d)(a+d)(b+d)}
=3(c+d){ab-cd-(a+d)(b+d)}
この式を

=3(c+d){ab-cd-(a+d)(b+d)}