The formula for the kth term of
(A + B)N is
C(N,k-1)AN-k+1Bk-1
The eighth term of the expansion (2c+d)^7
So substitute A = 2c, B = d, k = 8, N = 7
C(N,k-1)AN-k+1Bk-1
C(7,8-1)(2c)7-8+1d8-1
C(7,7)(2c)0d7
(1)(1)d7
d7
The first term of the expansion of (3q-7r)^6
So substitute A = 3q, B = -7r, k = 1, N = 6
C(N,k-1)AN-k+1Bk-1
C(6,1-1)(3q)6-1+1(-7r)1-1
C(6,0)(3q)6(-7r)0
(1)(3q)6(1)
(3q)6
36q6
729q6
Edwin