@MZH_Maths A²-A³ = 12 A³-A²+12 = 0 A³+2A²-3A²+12 = 0 A²(A+2)-3(A²-4) = 0 A²(A+2)-3(A-2)(A+2) = 0 (A+2)[A²-3(A-2)] = 0 (A+2)[A²-3A+6] = 0 A+2 = 0 A = -2 <== #answer 1A²-3A+6 = 0 X = 1 Y = -3 Z = 6 A = [3±√(9-4*1*6)] /2 = (½)(3±i√15) <== answers
@MZH_Maths A²-A³ = 12 A³-A²+12 = 0 A³+2A²-3A²+12 = 0 A²(A+2)-3(A²-4) = 0 A²(A+2)-3(A-2)(A+2) = 0 (A+2)[A²-3(A-2)] = 0 (A+2)[A²-3A+6] = 0 A+2 = 0 A = -2 <== #answer 1A²-3A+6 = 0 X = 1 Y = -3 Z = 6 A = [3±√(9-4*1*6)] /2 = (½)(3±i√15) <== answers