投稿

検索キーワード「whole cube a b c 3」に一致する投稿を表示しています

√100以上 (a+b+c)^3 proof 225102-A^3+b^3+c^3-3abc proof

イメージ
Math a^3 b^3 = (a b)(a^2 b^2 ab) /math Lets try to derive this expansion from the expansion of math (a b) ^ 3 /math We have, math(a b) ^ 3 = a^3A = 4 3 σ ( σ − m a ) ( σ − m b ) ( σ − m c ) {\displaystyle A= {\frac {4} {3}} {\sqrt {\sigma (\sigma m_ {a}) (\sigma m_ {b}) (\sigma m_ {c})}}} Next, denoting the altitudes from sides a, b, and c respectively as ha, hb, and hc, and denoting the semisum of the reciprocals of the altitudes as H = 1 / 2 (h−1Misc 3 Let A, B and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C show that B = C In order to prove B = C, we should prove B is a subset of C ie B ⊂ C & C is a subset of B ie C ⊂ B Let x ∈ B ⇒ x ∈ A ∪ B ⇒ x ∈ A ∪ C ⇒ x ∈ A or x ∈ C Sample Problems A^3+b^3+c^3-3abc proof