Metamath Proof Explorer


Theorem 3cubeslem3

Description: Lemma for 3cubes . (Contributed by Igor Ieskov, 22-Jan-2024)

Ref Expression
Hypothesis 3cubeslem1.a ⊢ φ → A ∈ ℚ
Assertion 3cubeslem3 ⊢ φ → A ⁢ 3 3 ⁢ A 2 + 3 2 ⁢ A + 3 3 = 3 3 ⁢ A 3 − 1 3 + − 3 3 ⁢ A 3 + 3 2 ⁢ A + 1 3 + 3 3 ⁢ A 2 + 3 2 ⁢ A 3

Proof

Step Hyp Ref Expression
1 3cubeslem1.a ⊢ φ → A ∈ ℚ
2 1 3cubeslem3l ⊢ φ → A ⁢ 3 3 ⁢ A 2 + 3 2 ⁢ A + 3 3 = A 7 ⁢ 3 9 + A 6 ⁢ 3 9 + A 5 ⁢ 3 8 + 3 8 + A 4 ⁢ 3 7 ⋅ 2 + 3 6 + A 3 ⁢ 3 6 + 3 6 + A 2 ⁢ 3 5 + A ⁢ 3 3
3 1 3cubeslem3r ⊢ φ → 3 3 ⁢ A 3 − 1 3 + − 3 3 ⁢ A 3 + 3 2 ⁢ A + 1 3 + 3 3 ⁢ A 2 + 3 2 ⁢ A 3 = A 7 ⁢ 3 9 + A 6 ⁢ 3 9 + A 5 ⁢ 3 8 + 3 8 + A 4 ⁢ 3 7 ⋅ 2 + 3 6 + A 3 ⁢ 3 6 + 3 6 + A 2 ⁢ 3 5 + A ⁢ 3 3
4 2 3 eqtr4d ⊢ φ → A ⁢ 3 3 ⁢ A 2 + 3 2 ⁢ A + 3 3 = 3 3 ⁢ A 3 − 1 3 + − 3 3 ⁢ A 3 + 3 2 ⁢ A + 1 3 + 3 3 ⁢ A 2 + 3 2 ⁢ A 3