Metamath Proof Explorer


Theorem dcubic1

Description: Forward direction of dcubic : the claimed formula produces solutions to the cubic equation. (Contributed by Mario Carneiro, 25-Apr-2015)

Ref Expression
Hypotheses dcubic.c ⊢ φ → P ∈ ℂ
dcubic.d ⊢ φ → Q ∈ ℂ
dcubic.x ⊢ φ → X ∈ ℂ
dcubic.t ⊢ φ → T ∈ ℂ
dcubic.3 ⊢ φ → T 3 = G − N
dcubic.g ⊢ φ → G ∈ ℂ
dcubic.2 ⊢ φ → G 2 = N 2 + M 3
dcubic.m ⊢ φ → M = P 3
dcubic.n ⊢ φ → N = Q 2
dcubic.0 ⊢ φ → T ≠ 0
dcubic1.x ⊢ φ → X = T − M T
Assertion dcubic1 ⊢ φ → X 3 + P ⁢ X + Q = 0

Proof

Step Hyp Ref Expression
1 dcubic.c ⊢ φ → P ∈ ℂ
2 dcubic.d ⊢ φ → Q ∈ ℂ
3 dcubic.x ⊢ φ → X ∈ ℂ
4 dcubic.t ⊢ φ → T ∈ ℂ
5 dcubic.3 ⊢ φ → T 3 = G − N
6 dcubic.g ⊢ φ → G ∈ ℂ
7 dcubic.2 ⊢ φ → G 2 = N 2 + M 3
8 dcubic.m ⊢ φ → M = P 3
9 dcubic.n ⊢ φ → N = Q 2
10 dcubic.0 ⊢ φ → T ≠ 0
11 dcubic1.x ⊢ φ → X = T − M T
12 5 oveq1d ⊢ φ → T 3 2 = G − N 2
13 2 halfcld ⊢ φ → Q 2 ∈ ℂ
14 9 13 eqeltrd ⊢ φ → N ∈ ℂ
15 binom2sub ⊢ G ∈ ℂ ∧ N ∈ ℂ → G − N 2 = G 2 - 2 ⁢ G ⋅ N + N 2
16 6 14 15 syl2anc ⊢ φ → G − N 2 = G 2 - 2 ⁢ G ⋅ N + N 2
17 2cnd ⊢ φ → 2 ∈ ℂ
18 17 6 14 mul12d ⊢ φ → 2 ⁢ G ⋅ N = G ⁢ 2 ⋅ N
19 9 oveq2d ⊢ φ → 2 ⋅ N = 2 ⁢ Q 2
20 2ne0 ⊢ 2 ≠ 0
21 20 a1i ⊢ φ → 2 ≠ 0
22 2 17 21 divcan2d ⊢ φ → 2 ⁢ Q 2 = Q
23 19 22 eqtrd ⊢ φ → 2 ⋅ N = Q
24 23 oveq2d ⊢ φ → G ⁢ 2 ⋅ N = G ⁢ Q
25 6 2 mulcomd ⊢ φ → G ⁢ Q = Q ⁢ G
26 18 24 25 3eqtrd ⊢ φ → 2 ⁢ G ⋅ N = Q ⁢ G
27 7 26 oveq12d ⊢ φ → G 2 − 2 ⁢ G ⋅ N = N 2 + M 3 - Q ⁢ G
28 27 oveq1d ⊢ φ → G 2 - 2 ⁢ G ⋅ N + N 2 = N 2 + M 3 - Q ⁢ G + N 2
29 12 16 28 3eqtrd ⊢ φ → T 3 2 = N 2 + M 3 - Q ⁢ G + N 2
30 14 sqcld ⊢ φ → N 2 ∈ ℂ
31 3cn ⊢ 3 ∈ ℂ
32 31 a1i ⊢ φ → 3 ∈ ℂ
33 3ne0 ⊢ 3 ≠ 0
34 33 a1i ⊢ φ → 3 ≠ 0
35 1 32 34 divcld ⊢ φ → P 3 ∈ ℂ
36 8 35 eqeltrd ⊢ φ → M ∈ ℂ
37 3nn0 ⊢ 3 ∈ ℕ 0
38 expcl ⊢ M ∈ ℂ ∧ 3 ∈ ℕ 0 → M 3 ∈ ℂ
39 36 37 38 sylancl ⊢ φ → M 3 ∈ ℂ
40 30 39 addcld ⊢ φ → N 2 + M 3 ∈ ℂ
41 2 6 mulcld ⊢ φ → Q ⁢ G ∈ ℂ
42 40 30 41 addsubd ⊢ φ → N 2 + M 3 + N 2 - Q ⁢ G = N 2 + M 3 - Q ⁢ G + N 2
43 30 39 30 add32d ⊢ φ → N 2 + M 3 + N 2 = N 2 + N 2 + M 3
44 30 2timesd ⊢ φ → 2 ⁢ N 2 = N 2 + N 2
45 44 oveq1d ⊢ φ → 2 ⁢ N 2 + M 3 = N 2 + N 2 + M 3
46 43 45 eqtr4d ⊢ φ → N 2 + M 3 + N 2 = 2 ⁢ N 2 + M 3
47 46 oveq1d ⊢ φ → N 2 + M 3 + N 2 - Q ⁢ G = 2 ⁢ N 2 + M 3 - Q ⁢ G
48 29 42 47 3eqtr2d ⊢ φ → T 3 2 = 2 ⁢ N 2 + M 3 - Q ⁢ G
49 2 6 14 subdid ⊢ φ → Q ⁢ G − N = Q ⁢ G − Q ⋅ N
50 5 oveq2d ⊢ φ → Q ⁢ T 3 = Q ⁢ G − N
51 14 sqvald ⊢ φ → N 2 = N ⋅ N
52 51 oveq2d ⊢ φ → 2 ⁢ N 2 = 2 ⁢ N ⋅ N
53 17 14 14 mulassd ⊢ φ → 2 ⋅ N ⋅ N = 2 ⁢ N ⋅ N
54 23 oveq1d ⊢ φ → 2 ⋅ N ⋅ N = Q ⋅ N
55 52 53 54 3eqtr2d ⊢ φ → 2 ⁢ N 2 = Q ⋅ N
56 55 oveq2d ⊢ φ → Q ⁢ G − 2 ⁢ N 2 = Q ⁢ G − Q ⋅ N
57 49 50 56 3eqtr4d ⊢ φ → Q ⁢ T 3 = Q ⁢ G − 2 ⁢ N 2
58 57 oveq1d ⊢ φ → Q ⁢ T 3 − M 3 = Q ⁢ G - 2 ⁢ N 2 - M 3
59 2cn ⊢ 2 ∈ ℂ
60 mulcl ⊢ 2 ∈ ℂ ∧ N 2 ∈ ℂ → 2 ⁢ N 2 ∈ ℂ
61 59 30 60 sylancr ⊢ φ → 2 ⁢ N 2 ∈ ℂ
62 41 61 39 subsub4d ⊢ φ → Q ⁢ G - 2 ⁢ N 2 - M 3 = Q ⁢ G − 2 ⁢ N 2 + M 3
63 58 62 eqtrd ⊢ φ → Q ⁢ T 3 − M 3 = Q ⁢ G − 2 ⁢ N 2 + M 3
64 48 63 oveq12d ⊢ φ → T 3 2 + Q ⁢ T 3 - M 3 = 2 ⁢ N 2 + M 3 - Q ⁢ G + Q ⁢ G - 2 ⁢ N 2 + M 3
65 61 39 addcld ⊢ φ → 2 ⁢ N 2 + M 3 ∈ ℂ
66 npncan2 ⊢ 2 ⁢ N 2 + M 3 ∈ ℂ ∧ Q ⁢ G ∈ ℂ → 2 ⁢ N 2 + M 3 - Q ⁢ G + Q ⁢ G - 2 ⁢ N 2 + M 3 = 0
67 65 41 66 syl2anc ⊢ φ → 2 ⁢ N 2 + M 3 - Q ⁢ G + Q ⁢ G - 2 ⁢ N 2 + M 3 = 0
68 64 67 eqtrd ⊢ φ → T 3 2 + Q ⁢ T 3 - M 3 = 0
69 1 2 3 4 5 6 7 8 9 10 4 10 11 dcubic1lem ⊢ φ → X 3 + P ⁢ X + Q = 0 ↔ T 3 2 + Q ⁢ T 3 - M 3 = 0
70 68 69 mpbird ⊢ φ → X 3 + P ⁢ X + Q = 0