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(a+b)³ + (b+c)³ + (a+c)³
= a³ + 3a²b + 3ab² + b³ + b³ + 3b²c + 3bc² + c³ + a³ + 3a²c + 3ac² + c³
= 2a³ + 2b³ + 2c³ + 3a²b + 3a²c + 3ab² + 3b²c + 3ac² + 3bc²
= 2(a³ + b³ + c³) + 3(a²b + a²c + ab² + b²c + ac² + bc²)
= 2((a+b+c)³ - (3(a²b + a²c + ab² + b²c + ac² + bc²) + 6abc) + 3(a²b + a²c + ab² + b²c + ac² + bc²)
= 2(a+b+c)³ - 6(a²b + a²c + ab² + b²c + ac² + bc²) + 3(a²b + a²c + ab² + b²c + ac² + bc²) - 12abc
= 2(a+b+c)³ - 3(a²b + a²c + ab² + b²c + ac² + bc²) - 12abc
= 2(a+b+c)³ - 3((a+b+c)(ab+ac+bc)-3abc) - 12abc
= 2(a+b+c)³ - 3(a+b+c)(ab+ac+bc) + 9abc - 12abc
= 2(a+b+c)³ - 3(a+b+c)(ab+ac+bc) - 3abc
Dengan teorema Vieta
a+b+c = - Koefisien x² / Koefisien x³ = - 0/5 = 0
ab+ac+bc = Koefisien x / Koefisien x³ = 2011/5
abc = - Konstanta / Koefisien x³ = -2015 / 5 = -403
Maka:
= 2(0)² - 3(0)(2011/5) - 3(-403)
= 1.209