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anonymous
 3 years ago
a+b+c=4
a^2+b^2+c^2=10
a^3+b^3+c^3=22
find the values of a, b ,and c
anonymous
 3 years ago
a+b+c=4 a^2+b^2+c^2=10 a^3+b^3+c^3=22 find the values of a, b ,and c

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I have no idea, I'm commenting so that I'll get a notification when someone solves this.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If someone gets this they are the smartest person I know!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0First note that:\[(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)\Longrightarrow \]\[4^2=10+2(ab+ac+bc)\Longrightarrow ab+ac+bc=3\]Also:\[(a+b+c)^3=a^3+b^3+c^3+3(a+b+c)(ab+ac+bc)3abc\Longrightarrow\]\[4^3=22+3(4)(3)3abc\Longrightarrow abc=2\]So your system is equivalent to: a+b+c=10 ab+ac+bc=3 abc=2 Solving this system is the same as solving the polynomial: \[x^310x^2+3x+2=0\]So the solutions to this polynomial are the solutions to your system.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh, mistype, thats why things were acting weird. Your system is equivalent to:\[a+b+c=4\]\[ab+ac+bc=3\]\[abc=2\]which gives the polynomial:\[x^34x+3x2=0\]which gives solutions \[a= 2, b=1\sqrt{2}, c=1+\sqrt{2}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0mistype again, +2 instead of 2 >.>

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You must have had loads of practice to be as good as you are!
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