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estudierBest ResponseYou've already chosen the best response.2
IF you can keep your head when all about you. Are losing theirs and blaming it on you, If you can trust yourself when all men doubt you, But make ...
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
\[a,b,c \]are roots of the equation\[x^3x1=0\] compute \[\frac{ 1a }{ 1+a }+\frac{ 1b }{ 1+b }+\frac{ 1c }{ 1+c}\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
simplified \[\frac{ 1 }{ a+1 }+\frac{ 1 }{ b+1 }+\frac{ 1 }{ c+1 }3\] note\[a^3=a+1,b^3=b+1,c^3=c+1\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
so\[\frac{ 1 }{ a^3 }+\frac{ 1 }{ b^3 }+\frac{ 1 }{ c^3 }3\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
if we divide by a,b,c\[1\frac{ 1 }{ a^2 }=\frac{ 1 }{ a^3 }\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
\[[3(\frac{ 1 }{ a^2 }+\frac{ 1 }{ b^2 }+\frac{ 1 }{ c^2 })]3\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
\[(xa)(xb)(xc)=x^3x^2(a+b+c)+x(bc+ab+ac)(abc)\] coefeecients of x^2 is 0 x is 1 and constant 0 so\[a+b+c=0,bc+ab+ac=1,abc=1\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
\[\frac{ (ac)^2+(bc)^2+(ab)^2}{ (abc)^2 }\] we need to get this with all the info
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
\[(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
\[(0)^2=(ab)^2+(bc)^2+(ca)^2+2(1)\] \[(ab)^2+(bc)^2+(ca)^2=2\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
so the answer to\[[3(\frac{ (ab)^2(bc)^2+(ca))^2 }{ (abc)^2 })]3=[3\frac{ 2 }{ 1 }]3=2\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
wat doyou think guys,please help
 one year ago

ZekariasBest ResponseYou've already chosen the best response.0
Actually the answer is 1 So you are not correct
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
how can we find the answer 1
 one year ago

ZekariasBest ResponseYou've already chosen the best response.0
\[\frac{ 1a }{ 1+a }+\frac{ 1b }{ 1+b }+\frac{ 1c }{ 1+c }\] \[1+\frac{ 2 }{ 1+a }1+\frac{ 2 }{ 1+b }1+\frac{ 2 }{ 1+c }\] \[2(\frac{ 1 }{ 1+a }+\frac{ 1 }{ 1+b }+\frac{ 1 }{ 1+c })3\] \[2(\frac{ (1+b)(1+c)+(1+a)(1+c)+(1+a)(1+b) }{ (1+a)(1+b)(1+c) })3\] \[2(\frac{ (ab+ac+bc)+2(a+b+c)+3 }{ abc+(ab+ac+bc)+(a+b+c)+1 })3\]Now insert the respective values so as to get the ANSWER 1
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
i am out of medals...thanks
 one year ago

estudierBest ResponseYou've already chosen the best response.2
Multiply it out directly: > 3 + (a + b + c)  (ab +bc + ca)  3abc / (a+1)(b+1)(c+1) > 3 + (a + b + c)  (ab +bc + ca)  3abc / a^3b^3c^3 = 3 +1  3 /1 = 1
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
i really went for miles to show a centimeter journey
 one year ago

estudierBest ResponseYou've already chosen the best response.2
Now and again, it is just a question of doing the first thing that comes to mind....:)
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
\[x^3−x−1=0\] 1 is not a solution as \[1^3−1−1=1\neq0\]
 one year ago

estudierBest ResponseYou've already chosen the best response.2
We are not finding solutions, we are evaluating the expression in the top post...
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
is it possible to getthe values ofa,b,c
 one year ago

estudierBest ResponseYou've already chosen the best response.2
http://en.wikipedia.org/wiki/Plastic_number
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
very helpful,i asked this becuause 1 was looking for the solution of the cube
 one year ago

estudierBest ResponseYou've already chosen the best response.2
http://en.wikipedia.org/wiki/Cubic_function#Reduction_to_a_depressed_cubic
 one year ago
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