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estudier Group TitleBest 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 ...
 2 years ago

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

Jonask Group TitleBest ResponseYou've already chosen the best response.1
simplified \[\frac{ 1 }{ a+1 }+\frac{ 1 }{ b+1 }+\frac{ 1 }{ c+1 }3\] note\[a^3=a+1,b^3=b+1,c^3=c+1\]
 2 years ago

Jonask Group TitleBest ResponseYou've already chosen the best response.1
so\[\frac{ 1 }{ a^3 }+\frac{ 1 }{ b^3 }+\frac{ 1 }{ c^3 }3\]
 2 years ago

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

Jonask Group TitleBest ResponseYou've already chosen the best response.1
\[[3(\frac{ 1 }{ a^2 }+\frac{ 1 }{ b^2 }+\frac{ 1 }{ c^2 })]3\]
 2 years ago

Jonask Group TitleBest 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\[a+b+c=0,bc+ab+ac=1,abc=1\]
 2 years ago

Jonask Group TitleBest 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
 2 years ago

Jonask Group TitleBest ResponseYou've already chosen the best response.1
\[(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)\]
 2 years ago

Jonask Group TitleBest ResponseYou've already chosen the best response.1
\[(0)^2=(ab)^2+(bc)^2+(ca)^2+2(1)\] \[(ab)^2+(bc)^2+(ca)^2=2\]
 2 years ago

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

Jonask Group TitleBest ResponseYou've already chosen the best response.1
wat doyou think guys,please help
 2 years ago

Zekarias Group TitleBest ResponseYou've already chosen the best response.0
Actually the answer is 1 So you are not correct
 2 years ago

Jonask Group TitleBest ResponseYou've already chosen the best response.1
how can we find the answer 1
 2 years ago

Zekarias Group TitleBest 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
 2 years ago

Jonask Group TitleBest ResponseYou've already chosen the best response.1
i am out of medals...thanks
 2 years ago

estudier Group TitleBest 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
 2 years ago

Jonask Group TitleBest ResponseYou've already chosen the best response.1
i really went for miles to show a centimeter journey
 2 years ago

estudier Group TitleBest 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....:)
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
\[x^3−x−1=0\] 1 is not a solution as \[1^3−1−1=1\neq0\]
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.2
We are not finding solutions, we are evaluating the expression in the top post...
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
to hard for me
 2 years ago

Jonask Group TitleBest ResponseYou've already chosen the best response.1
is it possible to getthe values ofa,b,c
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.2
Do you mean by hand?
 2 years ago

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

Jonask Group TitleBest 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

estudier Group TitleBest 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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