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SUPPLEMENTARY MATERIAL (USERS' MATHBOXES)
Mathbox for Igor Ieskov
cu3addd
Next ⟩
sqnegd
Metamath Proof Explorer
Ascii
Unicode
Theorem
cu3addd
Description:
Cube of sum of three numbers.
(Contributed by
Igor Ieskov
, 14-Dec-2023)
Ref
Expression
Hypotheses
cu3addd.1
⊢
φ
→
A
∈
ℂ
cu3addd.2
⊢
φ
→
B
∈
ℂ
cu3addd.3
⊢
φ
→
C
∈
ℂ
Assertion
cu3addd
⊢
φ
→
A
+
B
+
C
3
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
3
⋅
2
A
B
C
+
3
B
2
C
+
3
A
C
2
+
3
B
C
2
+
C
3
Proof
Step
Hyp
Ref
Expression
1
cu3addd.1
⊢
φ
→
A
∈
ℂ
2
cu3addd.2
⊢
φ
→
B
∈
ℂ
3
cu3addd.3
⊢
φ
→
C
∈
ℂ
4
1
2
addcld
⊢
φ
→
A
+
B
∈
ℂ
5
4
3
jca
⊢
φ
→
A
+
B
∈
ℂ
∧
C
∈
ℂ
6
binom3
⊢
A
+
B
∈
ℂ
∧
C
∈
ℂ
→
A
+
B
+
C
3
=
A
+
B
3
+
3
A
+
B
2
C
+
3
A
+
B
C
2
+
C
3
7
6
a1i
⊢
φ
→
A
+
B
∈
ℂ
∧
C
∈
ℂ
→
A
+
B
+
C
3
=
A
+
B
3
+
3
A
+
B
2
C
+
3
A
+
B
C
2
+
C
3
8
5
7
mpd
⊢
φ
→
A
+
B
+
C
3
=
A
+
B
3
+
3
A
+
B
2
C
+
3
A
+
B
C
2
+
C
3
9
binom3
⊢
A
∈
ℂ
∧
B
∈
ℂ
→
A
+
B
3
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
10
1
2
9
syl2anc
⊢
φ
→
A
+
B
3
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
11
10
oveq1d
⊢
φ
→
A
+
B
3
+
3
A
+
B
2
C
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
+
B
2
C
12
11
oveq1d
⊢
φ
→
A
+
B
3
+
3
A
+
B
2
C
+
3
A
+
B
C
2
+
C
3
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
+
B
2
C
+
3
A
+
B
C
2
+
C
3
13
8
12
eqtrd
⊢
φ
→
A
+
B
+
C
3
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
+
B
2
C
+
3
A
+
B
C
2
+
C
3
14
1
2
binom2d
⊢
φ
→
A
+
B
2
=
A
2
+
2
A
B
+
B
2
15
14
oveq1d
⊢
φ
→
A
+
B
2
C
=
A
2
+
2
A
B
+
B
2
C
16
15
oveq2d
⊢
φ
→
3
A
+
B
2
C
=
3
A
2
+
2
A
B
+
B
2
C
17
16
oveq2d
⊢
φ
→
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
+
B
2
C
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
+
2
A
B
+
B
2
C
18
17
oveq1d
⊢
φ
→
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
+
B
2
C
+
3
A
+
B
C
2
+
C
3
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
+
2
A
B
+
B
2
C
+
3
A
+
B
C
2
+
C
3
19
13
18
eqtrd
⊢
φ
→
A
+
B
+
C
3
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
+
2
A
B
+
B
2
C
+
3
A
+
B
C
2
+
C
3
20
1
sqcld
⊢
φ
→
A
2
∈
ℂ
21
2cnd
⊢
φ
→
2
∈
ℂ
22
1
2
mulcld
⊢
φ
→
A
B
∈
ℂ
23
21
22
mulcld
⊢
φ
→
2
A
B
∈
ℂ
24
20
23
addcld
⊢
φ
→
A
2
+
2
A
B
∈
ℂ
25
2
sqcld
⊢
φ
→
B
2
∈
ℂ
26
24
25
3
adddird
⊢
φ
→
A
2
+
2
A
B
+
B
2
C
=
A
2
+
2
A
B
C
+
B
2
C
27
26
oveq2d
⊢
φ
→
3
A
2
+
2
A
B
+
B
2
C
=
3
A
2
+
2
A
B
C
+
B
2
C
28
27
oveq2d
⊢
φ
→
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
+
2
A
B
+
B
2
C
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
+
2
A
B
C
+
B
2
C
29
28
oveq1d
⊢
φ
→
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
+
2
A
B
+
B
2
C
+
3
A
+
B
C
2
+
C
3
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
+
2
A
B
C
+
B
2
C
+
3
A
+
B
C
2
+
C
3
30
19
29
eqtrd
⊢
φ
→
A
+
B
+
C
3
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
+
2
A
B
C
+
B
2
C
+
3
A
+
B
C
2
+
C
3
31
20
23
3
adddird
⊢
φ
→
A
2
+
2
A
B
C
=
A
2
C
+
2
A
B
C
32
31
oveq1d
⊢
φ
→
A
2
+
2
A
B
C
+
B
2
C
=
A
2
C
+
2
A
B
C
+
B
2
C
33
32
oveq2d
⊢
φ
→
3
A
2
+
2
A
B
C
+
B
2
C
=
3
A
2
C
+
2
A
B
C
+
B
2
C
34
33
oveq2d
⊢
φ
→
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
+
2
A
B
C
+
B
2
C
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
2
A
B
C
+
B
2
C
35
34
oveq1d
⊢
φ
→
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
+
2
A
B
C
+
B
2
C
+
3
A
+
B
C
2
+
C
3
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
2
A
B
C
+
B
2
C
+
3
A
+
B
C
2
+
C
3
36
30
35
eqtrd
⊢
φ
→
A
+
B
+
C
3
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
2
A
B
C
+
B
2
C
+
3
A
+
B
C
2
+
C
3
37
3cn
⊢
3
∈
ℂ
38
37
a1i
⊢
φ
→
3
∈
ℂ
39
20
3
mulcld
⊢
φ
→
A
2
C
∈
ℂ
40
23
3
mulcld
⊢
φ
→
2
A
B
C
∈
ℂ
41
39
40
addcld
⊢
φ
→
A
2
C
+
2
A
B
C
∈
ℂ
42
25
3
mulcld
⊢
φ
→
B
2
C
∈
ℂ
43
38
41
42
adddid
⊢
φ
→
3
A
2
C
+
2
A
B
C
+
B
2
C
=
3
A
2
C
+
2
A
B
C
+
3
B
2
C
44
43
oveq2d
⊢
φ
→
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
2
A
B
C
+
B
2
C
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
2
A
B
C
+
3
B
2
C
45
44
oveq1d
⊢
φ
→
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
2
A
B
C
+
B
2
C
+
3
A
+
B
C
2
+
C
3
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
2
A
B
C
+
3
B
2
C
+
3
A
+
B
C
2
+
C
3
46
36
45
eqtrd
⊢
φ
→
A
+
B
+
C
3
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
2
A
B
C
+
3
B
2
C
+
3
A
+
B
C
2
+
C
3
47
38
39
40
adddid
⊢
φ
→
3
A
2
C
+
2
A
B
C
=
3
A
2
C
+
3
2
A
B
C
48
47
oveq1d
⊢
φ
→
3
A
2
C
+
2
A
B
C
+
3
B
2
C
=
3
A
2
C
+
3
2
A
B
C
+
3
B
2
C
49
48
oveq2d
⊢
φ
→
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
2
A
B
C
+
3
B
2
C
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
3
2
A
B
C
+
3
B
2
C
50
49
oveq1d
⊢
φ
→
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
2
A
B
C
+
3
B
2
C
+
3
A
+
B
C
2
+
C
3
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
3
2
A
B
C
+
3
B
2
C
+
3
A
+
B
C
2
+
C
3
51
46
50
eqtrd
⊢
φ
→
A
+
B
+
C
3
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
3
2
A
B
C
+
3
B
2
C
+
3
A
+
B
C
2
+
C
3
52
38
21
22
mulassd
⊢
φ
→
3
⋅
2
A
B
=
3
2
A
B
53
52
oveq1d
⊢
φ
→
3
⋅
2
A
B
C
=
3
2
A
B
C
54
38
23
3
mulassd
⊢
φ
→
3
2
A
B
C
=
3
2
A
B
C
55
53
54
eqtrd
⊢
φ
→
3
⋅
2
A
B
C
=
3
2
A
B
C
56
55
oveq2d
⊢
φ
→
3
A
2
C
+
3
⋅
2
A
B
C
=
3
A
2
C
+
3
2
A
B
C
57
56
oveq1d
⊢
φ
→
3
A
2
C
+
3
⋅
2
A
B
C
+
3
B
2
C
=
3
A
2
C
+
3
2
A
B
C
+
3
B
2
C
58
57
oveq2d
⊢
φ
→
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
3
⋅
2
A
B
C
+
3
B
2
C
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
3
2
A
B
C
+
3
B
2
C
59
58
eqcomd
⊢
φ
→
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
3
2
A
B
C
+
3
B
2
C
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
3
⋅
2
A
B
C
+
3
B
2
C
60
59
oveq1d
⊢
φ
→
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
3
2
A
B
C
+
3
B
2
C
+
3
A
+
B
C
2
+
C
3
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
3
⋅
2
A
B
C
+
3
B
2
C
+
3
A
+
B
C
2
+
C
3
61
51
60
eqtrd
⊢
φ
→
A
+
B
+
C
3
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
3
⋅
2
A
B
C
+
3
B
2
C
+
3
A
+
B
C
2
+
C
3
62
3
sqcld
⊢
φ
→
C
2
∈
ℂ
63
1
2
62
adddird
⊢
φ
→
A
+
B
C
2
=
A
C
2
+
B
C
2
64
63
oveq2d
⊢
φ
→
3
A
+
B
C
2
=
3
A
C
2
+
B
C
2
65
64
oveq1d
⊢
φ
→
3
A
+
B
C
2
+
C
3
=
3
A
C
2
+
B
C
2
+
C
3
66
65
oveq2d
⊢
φ
→
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
3
⋅
2
A
B
C
+
3
B
2
C
+
3
A
+
B
C
2
+
C
3
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
3
⋅
2
A
B
C
+
3
B
2
C
+
3
A
C
2
+
B
C
2
+
C
3
67
61
66
eqtrd
⊢
φ
→
A
+
B
+
C
3
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
3
⋅
2
A
B
C
+
3
B
2
C
+
3
A
C
2
+
B
C
2
+
C
3
68
1
62
mulcld
⊢
φ
→
A
C
2
∈
ℂ
69
2
62
mulcld
⊢
φ
→
B
C
2
∈
ℂ
70
38
68
69
adddid
⊢
φ
→
3
A
C
2
+
B
C
2
=
3
A
C
2
+
3
B
C
2
71
70
oveq1d
⊢
φ
→
3
A
C
2
+
B
C
2
+
C
3
=
3
A
C
2
+
3
B
C
2
+
C
3
72
71
oveq2d
⊢
φ
→
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
3
⋅
2
A
B
C
+
3
B
2
C
+
3
A
C
2
+
B
C
2
+
C
3
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
3
⋅
2
A
B
C
+
3
B
2
C
+
3
A
C
2
+
3
B
C
2
+
C
3
73
67
72
eqtrd
⊢
φ
→
A
+
B
+
C
3
=
A
3
+
3
A
2
B
+
3
A
B
2
+
B
3
+
3
A
2
C
+
3
⋅
2
A
B
C
+
3
B
2
C
+
3
A
C
2
+
3
B
C
2
+
C
3