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(ITA-79) Se a, b, c são as raízes da equação x³-rx+20=0 onde r é um número real, podemos afirmar que o valor de a³+b³+c³ é?
2 Answers
- EinsteinLv 74 years ago
x³ + 0x² - rx + 20 = 0 ---> Raízes: a, b, c
x³-rx+20=(x-a)(x-b)(x-c)
x³-rx+20=(x²-x(a+b)+ab)(x-c)
x³-rx+20=x³-x²(a+b)+abx-cx²+x(ac+bc)-abc
x³-rx+20=x³-x²(a+b+c)+x(ac+bc+ab)-abc
Girard:
a + b + c = 0 ----> I
ab + ac + bc = - r ----> II
abc = - 20 ----> III
I*II ----> (a + b + c)*(ab + ac + bc) = 0*(-r) ----> a²b + a²c + ab² + cb² + ac² + bc² = - 3abc ----> IV
I ----> (a + b + c)³ = 0³----> a³ + b³ + c² + 3*(a²b + a²c + ab² + cb² + ac² + bc²) + 6abc = 0 ----> V
IV em V ----> a³ + b³ + c³ + 3*(- 3abc) + 6abc = 0 ----> a³ + b³ + c³ = 3abc ----> a³ + b³ + c³ = - 60