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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
 one year ago
 one year 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
 one year ago
 one year ago

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liza2Best ResponseYou've already chosen the best response.0
I have no idea, I'm commenting so that I'll get a notification when someone solves this.
 one year ago

alexd00dBest ResponseYou've already chosen the best response.0
If someone gets this they are the smartest person I know!
 one year ago

joemath314159Best ResponseYou've already chosen the best response.1
First 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.
 one year ago

joemath314159Best ResponseYou've already chosen the best response.1
oh, 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 \[a= 2, b=1\sqrt{2}, c=1+\sqrt{2}\]
 one year ago

joemath314159Best ResponseYou've already chosen the best response.1
mistype again, +2 instead of 2 >.>
 one year ago

alexd00dBest ResponseYou've already chosen the best response.0
You must have had loads of practice to be as good as you are!
 one year ago
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