a+b+c = 1 a²+b²+c² = 9 a³+b³+c³ = 1
Find 1/a + 1/b + 1/c
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a+b+c = 1 a²+b²+c² = 9 a³+b³+c³ = 1
Find 1/a + 1/b + 1/c
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Answer:
answer. 1
Step-by-step explanation:
[tex]a + b + c = 1 \\ \\ {a}^{2} + {b}^{2} + {c }^{2} = 9 \\ \\ {a}^{3} + {b}^{3} + {c}^{3} = 1 \\ \\ {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca \\ {1}^{2} = 9 + 2(ab + bc + ca) \\ 1 - 9 = 2(ab + bc + ca) \\ \\ ab + bc + ca = - \frac{8}{2} = - 4 \\ \\ \\ \\ {a}^{3} + {b}^{3} + {c}^{3} - 3abc = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca) \\ \\ 1 - 3abc = 1(9 - ( - 4)) \\ \\ 1 - 3abc = 13 \\ - 3abc = 13 - 1 \\ \\ - 3abc = 12 \\ \\ abc = - 4 \\ \\ now \: \: \: \: \: \: \frac{ ab + bc + ca }{abc} = \frac{ - 4}{ - 4} \\ \\ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1[/tex]