Metamath Proof Explorer


Theorem binom4

Description: Work out a quartic binomial. (You would think that by this point it would be faster to use binom , but it turns out to be just as much work to put it into this form after clearing all the sums and calculating binomial coefficients.) (Contributed by Mario Carneiro, 6-May-2015)

Ref Expression
Assertion binom4 ⊢ A ∈ ℂ ∧ B ∈ ℂ → A + B 4 = A 4 + 4 ⁢ A 3 ⁢ B + 6 ⁢ A 2 ⁢ B 2 + 4 ⁢ A ⁢ B 3 + B 4

Proof

Step Hyp Ref Expression
1 df-4 ⊢ 4 = 3 + 1
2 1 oveq2i ⊢ A + B 4 = A + B 3 + 1
3 addcl ⊢ A ∈ ℂ ∧ B ∈ ℂ → A + B ∈ ℂ
4 3nn0 ⊢ 3 ∈ ℕ 0
5 expp1 ⊢ A + B ∈ ℂ ∧ 3 ∈ ℕ 0 → A + B 3 + 1 = A + B 3 ⁢ A + B
6 3 4 5 sylancl ⊢ A ∈ ℂ ∧ B ∈ ℂ → A + B 3 + 1 = A + B 3 ⁢ A + B
7 2 6 eqtrid ⊢ A ∈ ℂ ∧ B ∈ ℂ → A + B 4 = A + B 3 ⁢ A + B
8 binom3 ⊢ A ∈ ℂ ∧ B ∈ ℂ → A + B 3 = A 3 + 3 ⁢ A 2 ⁢ B + 3 ⁢ A ⁢ B 2 + B 3
9 8 oveq1d ⊢ A ∈ ℂ ∧ B ∈ ℂ → A + B 3 ⁢ A + B = A 3 + 3 ⁢ A 2 ⁢ B + 3 ⁢ A ⁢ B 2 + B 3 ⁢ A + B
10 simpl ⊢ A ∈ ℂ ∧ B ∈ ℂ → A ∈ ℂ
11 expcl ⊢ A ∈ ℂ ∧ 3 ∈ ℕ 0 → A 3 ∈ ℂ
12 10 4 11 sylancl ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 3 ∈ ℂ
13 3cn ⊢ 3 ∈ ℂ
14 10 sqcld ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 2 ∈ ℂ
15 simpr ⊢ A ∈ ℂ ∧ B ∈ ℂ → B ∈ ℂ
16 14 15 mulcld ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 2 ⁢ B ∈ ℂ
17 mulcl ⊢ 3 ∈ ℂ ∧ A 2 ⁢ B ∈ ℂ → 3 ⁢ A 2 ⁢ B ∈ ℂ
18 13 16 17 sylancr ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A 2 ⁢ B ∈ ℂ
19 12 18 addcld ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 3 + 3 ⁢ A 2 ⁢ B ∈ ℂ
20 15 sqcld ⊢ A ∈ ℂ ∧ B ∈ ℂ → B 2 ∈ ℂ
21 10 20 mulcld ⊢ A ∈ ℂ ∧ B ∈ ℂ → A ⁢ B 2 ∈ ℂ
22 mulcl ⊢ 3 ∈ ℂ ∧ A ⁢ B 2 ∈ ℂ → 3 ⁢ A ⁢ B 2 ∈ ℂ
23 13 21 22 sylancr ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A ⁢ B 2 ∈ ℂ
24 expcl ⊢ B ∈ ℂ ∧ 3 ∈ ℕ 0 → B 3 ∈ ℂ
25 15 4 24 sylancl ⊢ A ∈ ℂ ∧ B ∈ ℂ → B 3 ∈ ℂ
26 23 25 addcld ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A ⁢ B 2 + B 3 ∈ ℂ
27 19 26 addcld ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 3 + 3 ⁢ A 2 ⁢ B + 3 ⁢ A ⁢ B 2 + B 3 ∈ ℂ
28 27 10 15 adddid ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 3 + 3 ⁢ A 2 ⁢ B + 3 ⁢ A ⁢ B 2 + B 3 ⁢ A + B = A 3 + 3 ⁢ A 2 ⁢ B + 3 ⁢ A ⁢ B 2 + B 3 ⁢ A + A 3 + 3 ⁢ A 2 ⁢ B + 3 ⁢ A ⁢ B 2 + B 3 ⁢ B
29 1 oveq2i ⊢ A 4 = A 3 + 1
30 expp1 ⊢ A ∈ ℂ ∧ 3 ∈ ℕ 0 → A 3 + 1 = A 3 ⁢ A
31 10 4 30 sylancl ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 3 + 1 = A 3 ⁢ A
32 29 31 eqtr2id ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 3 ⁢ A = A 4
33 13 a1i ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ∈ ℂ
34 33 16 10 mulassd ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A 2 ⁢ B ⁢ A = 3 ⁢ A 2 ⁢ B ⁢ A
35 14 15 10 mul32d ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 2 ⁢ B ⁢ A = A 2 ⁢ A ⁢ B
36 df-3 ⊢ 3 = 2 + 1
37 36 oveq2i ⊢ A 3 = A 2 + 1
38 2nn0 ⊢ 2 ∈ ℕ 0
39 expp1 ⊢ A ∈ ℂ ∧ 2 ∈ ℕ 0 → A 2 + 1 = A 2 ⁢ A
40 10 38 39 sylancl ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 2 + 1 = A 2 ⁢ A
41 37 40 eqtr2id ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 2 ⁢ A = A 3
42 41 oveq1d ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 2 ⁢ A ⁢ B = A 3 ⁢ B
43 35 42 eqtrd ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 2 ⁢ B ⁢ A = A 3 ⁢ B
44 43 oveq2d ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A 2 ⁢ B ⁢ A = 3 ⁢ A 3 ⁢ B
45 34 44 eqtrd ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A 2 ⁢ B ⁢ A = 3 ⁢ A 3 ⁢ B
46 32 45 oveq12d ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 3 ⁢ A + 3 ⁢ A 2 ⁢ B ⁢ A = A 4 + 3 ⁢ A 3 ⁢ B
47 12 10 18 46 joinlmuladdmuld ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 3 + 3 ⁢ A 2 ⁢ B ⁢ A = A 4 + 3 ⁢ A 3 ⁢ B
48 33 21 10 mulassd ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A ⁢ B 2 ⁢ A = 3 ⁢ A ⁢ B 2 ⁢ A
49 10 20 10 mul32d ⊢ A ∈ ℂ ∧ B ∈ ℂ → A ⁢ B 2 ⁢ A = A ⁢ A ⁢ B 2
50 10 sqvald ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 2 = A ⁢ A
51 50 oveq1d ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 2 ⁢ B 2 = A ⁢ A ⁢ B 2
52 49 51 eqtr4d ⊢ A ∈ ℂ ∧ B ∈ ℂ → A ⁢ B 2 ⁢ A = A 2 ⁢ B 2
53 52 oveq2d ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A ⁢ B 2 ⁢ A = 3 ⁢ A 2 ⁢ B 2
54 48 53 eqtrd ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A ⁢ B 2 ⁢ A = 3 ⁢ A 2 ⁢ B 2
55 25 10 mulcomd ⊢ A ∈ ℂ ∧ B ∈ ℂ → B 3 ⁢ A = A ⁢ B 3
56 54 55 oveq12d ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A ⁢ B 2 ⁢ A + B 3 ⁢ A = 3 ⁢ A 2 ⁢ B 2 + A ⁢ B 3
57 23 10 25 56 joinlmuladdmuld ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A ⁢ B 2 + B 3 ⁢ A = 3 ⁢ A 2 ⁢ B 2 + A ⁢ B 3
58 47 57 oveq12d ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 3 + 3 ⁢ A 2 ⁢ B ⁢ A + 3 ⁢ A ⁢ B 2 + B 3 ⁢ A = A 4 + 3 ⁢ A 3 ⁢ B + 3 ⁢ A 2 ⁢ B 2 + A ⁢ B 3
59 19 10 26 58 joinlmuladdmuld ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 3 + 3 ⁢ A 2 ⁢ B + 3 ⁢ A ⁢ B 2 + B 3 ⁢ A = A 4 + 3 ⁢ A 3 ⁢ B + 3 ⁢ A 2 ⁢ B 2 + A ⁢ B 3
60 19 26 15 adddird ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 3 + 3 ⁢ A 2 ⁢ B + 3 ⁢ A ⁢ B 2 + B 3 ⁢ B = A 3 + 3 ⁢ A 2 ⁢ B ⁢ B + 3 ⁢ A ⁢ B 2 + B 3 ⁢ B
61 33 16 15 mulassd ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A 2 ⁢ B ⁢ B = 3 ⁢ A 2 ⁢ B ⁢ B
62 14 15 15 mulassd ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 2 ⁢ B ⁢ B = A 2 ⁢ B ⁢ B
63 15 sqvald ⊢ A ∈ ℂ ∧ B ∈ ℂ → B 2 = B ⁢ B
64 63 oveq2d ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 2 ⁢ B 2 = A 2 ⁢ B ⁢ B
65 62 64 eqtr4d ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 2 ⁢ B ⁢ B = A 2 ⁢ B 2
66 65 oveq2d ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A 2 ⁢ B ⁢ B = 3 ⁢ A 2 ⁢ B 2
67 61 66 eqtrd ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A 2 ⁢ B ⁢ B = 3 ⁢ A 2 ⁢ B 2
68 67 oveq2d ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 3 ⁢ B + 3 ⁢ A 2 ⁢ B ⁢ B = A 3 ⁢ B + 3 ⁢ A 2 ⁢ B 2
69 12 15 18 68 joinlmuladdmuld ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 3 + 3 ⁢ A 2 ⁢ B ⁢ B = A 3 ⁢ B + 3 ⁢ A 2 ⁢ B 2
70 33 21 15 mulassd ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A ⁢ B 2 ⁢ B = 3 ⁢ A ⁢ B 2 ⁢ B
71 10 20 15 mulassd ⊢ A ∈ ℂ ∧ B ∈ ℂ → A ⁢ B 2 ⁢ B = A ⁢ B 2 ⁢ B
72 36 oveq2i ⊢ B 3 = B 2 + 1
73 expp1 ⊢ B ∈ ℂ ∧ 2 ∈ ℕ 0 → B 2 + 1 = B 2 ⁢ B
74 15 38 73 sylancl ⊢ A ∈ ℂ ∧ B ∈ ℂ → B 2 + 1 = B 2 ⁢ B
75 72 74 eqtr2id ⊢ A ∈ ℂ ∧ B ∈ ℂ → B 2 ⁢ B = B 3
76 75 oveq2d ⊢ A ∈ ℂ ∧ B ∈ ℂ → A ⁢ B 2 ⁢ B = A ⁢ B 3
77 71 76 eqtrd ⊢ A ∈ ℂ ∧ B ∈ ℂ → A ⁢ B 2 ⁢ B = A ⁢ B 3
78 77 oveq2d ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A ⁢ B 2 ⁢ B = 3 ⁢ A ⁢ B 3
79 70 78 eqtrd ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A ⁢ B 2 ⁢ B = 3 ⁢ A ⁢ B 3
80 1 oveq2i ⊢ B 4 = B 3 + 1
81 expp1 ⊢ B ∈ ℂ ∧ 3 ∈ ℕ 0 → B 3 + 1 = B 3 ⁢ B
82 15 4 81 sylancl ⊢ A ∈ ℂ ∧ B ∈ ℂ → B 3 + 1 = B 3 ⁢ B
83 80 82 eqtr2id ⊢ A ∈ ℂ ∧ B ∈ ℂ → B 3 ⁢ B = B 4
84 79 83 oveq12d ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A ⁢ B 2 ⁢ B + B 3 ⁢ B = 3 ⁢ A ⁢ B 3 + B 4
85 23 15 25 84 joinlmuladdmuld ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A ⁢ B 2 + B 3 ⁢ B = 3 ⁢ A ⁢ B 3 + B 4
86 69 85 oveq12d ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 3 + 3 ⁢ A 2 ⁢ B ⁢ B + 3 ⁢ A ⁢ B 2 + B 3 ⁢ B = A 3 ⁢ B + 3 ⁢ A 2 ⁢ B 2 + 3 ⁢ A ⁢ B 3 + B 4
87 12 15 mulcld ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 3 ⁢ B ∈ ℂ
88 14 20 mulcld ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 2 ⁢ B 2 ∈ ℂ
89 mulcl ⊢ 3 ∈ ℂ ∧ A 2 ⁢ B 2 ∈ ℂ → 3 ⁢ A 2 ⁢ B 2 ∈ ℂ
90 13 88 89 sylancr ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A 2 ⁢ B 2 ∈ ℂ
91 10 25 mulcld ⊢ A ∈ ℂ ∧ B ∈ ℂ → A ⁢ B 3 ∈ ℂ
92 mulcl ⊢ 3 ∈ ℂ ∧ A ⁢ B 3 ∈ ℂ → 3 ⁢ A ⁢ B 3 ∈ ℂ
93 13 91 92 sylancr ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A ⁢ B 3 ∈ ℂ
94 4nn0 ⊢ 4 ∈ ℕ 0
95 expcl ⊢ B ∈ ℂ ∧ 4 ∈ ℕ 0 → B 4 ∈ ℂ
96 15 94 95 sylancl ⊢ A ∈ ℂ ∧ B ∈ ℂ → B 4 ∈ ℂ
97 93 96 addcld ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A ⁢ B 3 + B 4 ∈ ℂ
98 87 90 97 addassd ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 3 ⁢ B + 3 ⁢ A 2 ⁢ B 2 + 3 ⁢ A ⁢ B 3 + B 4 = A 3 ⁢ B + 3 ⁢ A 2 ⁢ B 2 + 3 ⁢ A ⁢ B 3 + B 4
99 60 86 98 3eqtrd ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 3 + 3 ⁢ A 2 ⁢ B + 3 ⁢ A ⁢ B 2 + B 3 ⁢ B = A 3 ⁢ B + 3 ⁢ A 2 ⁢ B 2 + 3 ⁢ A ⁢ B 3 + B 4
100 59 99 oveq12d ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 3 + 3 ⁢ A 2 ⁢ B + 3 ⁢ A ⁢ B 2 + B 3 ⁢ A + A 3 + 3 ⁢ A 2 ⁢ B + 3 ⁢ A ⁢ B 2 + B 3 ⁢ B = A 4 + 3 ⁢ A 3 ⁢ B + 3 ⁢ A 2 ⁢ B 2 + A ⁢ B 3 + A 3 ⁢ B + 3 ⁢ A 2 ⁢ B 2 + 3 ⁢ A ⁢ B 3 + B 4
101 expcl ⊢ A ∈ ℂ ∧ 4 ∈ ℕ 0 → A 4 ∈ ℂ
102 10 94 101 sylancl ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 4 ∈ ℂ
103 mulcl ⊢ 3 ∈ ℂ ∧ A 3 ⁢ B ∈ ℂ → 3 ⁢ A 3 ⁢ B ∈ ℂ
104 13 87 103 sylancr ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A 3 ⁢ B ∈ ℂ
105 102 104 addcld ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 4 + 3 ⁢ A 3 ⁢ B ∈ ℂ
106 90 91 addcld ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A 2 ⁢ B 2 + A ⁢ B 3 ∈ ℂ
107 90 97 addcld ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A 2 ⁢ B 2 + 3 ⁢ A ⁢ B 3 + B 4 ∈ ℂ
108 105 106 87 107 add4d ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 4 + 3 ⁢ A 3 ⁢ B + 3 ⁢ A 2 ⁢ B 2 + A ⁢ B 3 + A 3 ⁢ B + 3 ⁢ A 2 ⁢ B 2 + 3 ⁢ A ⁢ B 3 + B 4 = A 4 + 3 ⁢ A 3 ⁢ B + A 3 ⁢ B + 3 ⁢ A 2 ⁢ B 2 + A ⁢ B 3 + 3 ⁢ A 2 ⁢ B 2 + 3 ⁢ A ⁢ B 3 + B 4
109 102 104 87 addassd ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 4 + 3 ⁢ A 3 ⁢ B + A 3 ⁢ B = A 4 + 3 ⁢ A 3 ⁢ B + A 3 ⁢ B
110 1 oveq1i ⊢ 4 ⁢ A 3 ⁢ B = 3 + 1 ⁢ A 3 ⁢ B
111 ax-1cn ⊢ 1 ∈ ℂ
112 111 a1i ⊢ A ∈ ℂ ∧ B ∈ ℂ → 1 ∈ ℂ
113 33 112 87 adddird ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 + 1 ⁢ A 3 ⁢ B = 3 ⁢ A 3 ⁢ B + 1 ⁢ A 3 ⁢ B
114 110 113 eqtrid ⊢ A ∈ ℂ ∧ B ∈ ℂ → 4 ⁢ A 3 ⁢ B = 3 ⁢ A 3 ⁢ B + 1 ⁢ A 3 ⁢ B
115 87 mullidd ⊢ A ∈ ℂ ∧ B ∈ ℂ → 1 ⁢ A 3 ⁢ B = A 3 ⁢ B
116 115 oveq2d ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A 3 ⁢ B + 1 ⁢ A 3 ⁢ B = 3 ⁢ A 3 ⁢ B + A 3 ⁢ B
117 114 116 eqtrd ⊢ A ∈ ℂ ∧ B ∈ ℂ → 4 ⁢ A 3 ⁢ B = 3 ⁢ A 3 ⁢ B + A 3 ⁢ B
118 117 oveq2d ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 4 + 4 ⁢ A 3 ⁢ B = A 4 + 3 ⁢ A 3 ⁢ B + A 3 ⁢ B
119 109 118 eqtr4d ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 4 + 3 ⁢ A 3 ⁢ B + A 3 ⁢ B = A 4 + 4 ⁢ A 3 ⁢ B
120 90 91 90 97 add4d ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A 2 ⁢ B 2 + A ⁢ B 3 + 3 ⁢ A 2 ⁢ B 2 + 3 ⁢ A ⁢ B 3 + B 4 = 3 ⁢ A 2 ⁢ B 2 + 3 ⁢ A 2 ⁢ B 2 + A ⁢ B 3 + 3 ⁢ A ⁢ B 3 + B 4
121 3p3e6 ⊢ 3 + 3 = 6
122 121 oveq1i ⊢ 3 + 3 ⁢ A 2 ⁢ B 2 = 6 ⁢ A 2 ⁢ B 2
123 33 33 88 adddird ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 + 3 ⁢ A 2 ⁢ B 2 = 3 ⁢ A 2 ⁢ B 2 + 3 ⁢ A 2 ⁢ B 2
124 122 123 eqtr3id ⊢ A ∈ ℂ ∧ B ∈ ℂ → 6 ⁢ A 2 ⁢ B 2 = 3 ⁢ A 2 ⁢ B 2 + 3 ⁢ A 2 ⁢ B 2
125 3p1e4 ⊢ 3 + 1 = 4
126 13 111 125 addcomli ⊢ 1 + 3 = 4
127 126 oveq1i ⊢ 1 + 3 ⁢ A ⁢ B 3 = 4 ⁢ A ⁢ B 3
128 112 33 91 adddird ⊢ A ∈ ℂ ∧ B ∈ ℂ → 1 + 3 ⁢ A ⁢ B 3 = 1 ⁢ A ⁢ B 3 + 3 ⁢ A ⁢ B 3
129 127 128 eqtr3id ⊢ A ∈ ℂ ∧ B ∈ ℂ → 4 ⁢ A ⁢ B 3 = 1 ⁢ A ⁢ B 3 + 3 ⁢ A ⁢ B 3
130 91 mullidd ⊢ A ∈ ℂ ∧ B ∈ ℂ → 1 ⁢ A ⁢ B 3 = A ⁢ B 3
131 130 oveq1d ⊢ A ∈ ℂ ∧ B ∈ ℂ → 1 ⁢ A ⁢ B 3 + 3 ⁢ A ⁢ B 3 = A ⁢ B 3 + 3 ⁢ A ⁢ B 3
132 129 131 eqtrd ⊢ A ∈ ℂ ∧ B ∈ ℂ → 4 ⁢ A ⁢ B 3 = A ⁢ B 3 + 3 ⁢ A ⁢ B 3
133 132 oveq1d ⊢ A ∈ ℂ ∧ B ∈ ℂ → 4 ⁢ A ⁢ B 3 + B 4 = A ⁢ B 3 + 3 ⁢ A ⁢ B 3 + B 4
134 91 93 96 addassd ⊢ A ∈ ℂ ∧ B ∈ ℂ → A ⁢ B 3 + 3 ⁢ A ⁢ B 3 + B 4 = A ⁢ B 3 + 3 ⁢ A ⁢ B 3 + B 4
135 133 134 eqtrd ⊢ A ∈ ℂ ∧ B ∈ ℂ → 4 ⁢ A ⁢ B 3 + B 4 = A ⁢ B 3 + 3 ⁢ A ⁢ B 3 + B 4
136 124 135 oveq12d ⊢ A ∈ ℂ ∧ B ∈ ℂ → 6 ⁢ A 2 ⁢ B 2 + 4 ⁢ A ⁢ B 3 + B 4 = 3 ⁢ A 2 ⁢ B 2 + 3 ⁢ A 2 ⁢ B 2 + A ⁢ B 3 + 3 ⁢ A ⁢ B 3 + B 4
137 120 136 eqtr4d ⊢ A ∈ ℂ ∧ B ∈ ℂ → 3 ⁢ A 2 ⁢ B 2 + A ⁢ B 3 + 3 ⁢ A 2 ⁢ B 2 + 3 ⁢ A ⁢ B 3 + B 4 = 6 ⁢ A 2 ⁢ B 2 + 4 ⁢ A ⁢ B 3 + B 4
138 119 137 oveq12d ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 4 + 3 ⁢ A 3 ⁢ B + A 3 ⁢ B + 3 ⁢ A 2 ⁢ B 2 + A ⁢ B 3 + 3 ⁢ A 2 ⁢ B 2 + 3 ⁢ A ⁢ B 3 + B 4 = A 4 + 4 ⁢ A 3 ⁢ B + 6 ⁢ A 2 ⁢ B 2 + 4 ⁢ A ⁢ B 3 + B 4
139 108 138 eqtrd ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 4 + 3 ⁢ A 3 ⁢ B + 3 ⁢ A 2 ⁢ B 2 + A ⁢ B 3 + A 3 ⁢ B + 3 ⁢ A 2 ⁢ B 2 + 3 ⁢ A ⁢ B 3 + B 4 = A 4 + 4 ⁢ A 3 ⁢ B + 6 ⁢ A 2 ⁢ B 2 + 4 ⁢ A ⁢ B 3 + B 4
140 28 100 139 3eqtrd ⊢ A ∈ ℂ ∧ B ∈ ℂ → A 3 + 3 ⁢ A 2 ⁢ B + 3 ⁢ A ⁢ B 2 + B 3 ⁢ A + B = A 4 + 4 ⁢ A 3 ⁢ B + 6 ⁢ A 2 ⁢ B 2 + 4 ⁢ A ⁢ B 3 + B 4
141 7 9 140 3eqtrd ⊢ A ∈ ℂ ∧ B ∈ ℂ → A + B 4 = A 4 + 4 ⁢ A 3 ⁢ B + 6 ⁢ A 2 ⁢ B 2 + 4 ⁢ A ⁢ B 3 + B 4