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estudier
 2 years ago
Best ResponseYou've already chosen the best response.2IF 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 ...

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

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

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

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

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

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

Jonask
 2 years ago
Best 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

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

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

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

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

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

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

Zekarias
 2 years ago
Best 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

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

estudier
 2 years ago
Best ResponseYou've already chosen the best response.2Multiply 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

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

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

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

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

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

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.1very helpful,i asked this becuause 1 was looking for the solution of the cube

estudier
 2 years ago
Best ResponseYou've already chosen the best response.2http://en.wikipedia.org/wiki/Cubic_function#Reduction_to_a_depressed_cubic
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