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検索キーワード「a3+b3」に一致する投稿を表示しています

(a-b)^3 formula proof 335773-A^3+b^3+c^3-3abc formula proof

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(ab) 2 = a 2 2ab b 2 (ab)(cd) = ac ad bc bd a 2 b 2 = (ab)(ab) (Difference of squares) a 3 b 3 = (a b)(a 2 ab b 2) (Sum and Difference of Cubes) x 2 (ab)x AB = (x a)(x b) if ax 2 bx c = 0 then x = ( b (b 2 4ac) ) / 2a (Quadratic Formula)And the quadratic formula was x The solutions would be equal to negative b plus or minus the square root of b squared minus 4ac, all of that over 2a And we learned how to use it You literally just substitute the numbers a for a, b for b, c for c, and then it gives you two answers, because you have a plus or a minus right thereA^3 – b^3 = (a – b)(a^2 ab b^2) a^3 b^3 = (a b)(a^2 – ab b^2) (a b)^3 = a^3 3a^2b 3ab^2 b^3 (a – b)^3 = a^3 – 3a^2b 3ab^2 – b^3 Binomial Theorem A^3+b^3+c^3-3abc formula proof