Answer
$((c+d)^2+1)(c+d+1)(c+d-1)$
Work Step by Step
The product of a binomial sum and a binomial difference is: $(A+B)(A-B)=A^2-B^2$.
The square of a binomial sum is: $(A+B)^2=A^2+2AB+B^2$.
The square of a binomial difference is: $(A-B)^2=A^2-2AB+B^2$.
Hence:
$(c+d)^4-1=\\=((c+d)^2)^2-1^2\\=((c+d)^2+1)((c+d)^2-1^2)\\=((c+d)^2+1)(c+d+1)(c+d-1)$
