Solution 1 Let f(A;B;C) = A3 C3 3ABC By direct computation, for nonnegative integers Awe have f(A;A;A 1) = 3A 1 and f(A;A;A) = 0, while for positive integers Awe have f(A;A;A 1) = 3A 1 and f(A;A 1;A 1) = 9A This shows that all the values listed in the answer can actually be obtained ToEQUIVALENCIAS CONDICION CERO Sea A B C = 0 • A 2 B 2 C 2 = 2(AB AC BC) • A 3 B 3 C 3 = 3ABC • A 4 B 4 C 4 = 2 2)2C2B2A( =2(A 2 B 2 A 2 C 2 B 2 C 2 ) EN CLASE 1 Sabiendo que ab = 5 y ab = 3 Calcula el valor de a² b² 2 Calcula 8 84 )15)(15)(1²5(241 3 SI (2xyz)² (2xyz)² = ²)²2(2 zxyWe know that a 3 b 3 c 3 – 3abc = (a b c) {a 2 b 2 c 2 –(ab bc ca)} ∴ a 3 b 3 c 3 – 3abc = 8 × (24 – ) = 4 × 8 = 32 ∴ a b c = 8, ab bc ca = and a 2 b 2 c 2 = 24 Thus, a 3 b 3 c 3 – 3abc = 32 TYPE OF FACTORIZATION 1) Factorization by taking out the common factors Ex ab(a 2 b 2 – c 2) bc (a 2 b 2 – c 2) – ca(a 2 b 2 – c 2) Sol A 3 B 3 C 3 3abc A B C A 2 B 2 C 2 Ab Ca Herunterladen A^3+b^3+c^3-3abc all formula
