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anonymous
 4 years ago
if
anonymous
 4 years ago
if

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0IF 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 ...

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0simplified \[\frac{ 1 }{ a+1 }+\frac{ 1 }{ b+1 }+\frac{ 1 }{ c+1 }3\] note\[a^3=a+1,b^3=b+1,c^3=c+1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so\[\frac{ 1 }{ a^3 }+\frac{ 1 }{ b^3 }+\frac{ 1 }{ c^3 }3\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if we divide by a,b,c\[1\frac{ 1 }{ a^2 }=\frac{ 1 }{ a^3 }\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[[3(\frac{ 1 }{ a^2 }+\frac{ 1 }{ b^2 }+\frac{ 1 }{ c^2 })]3\]

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ (ac)^2+(bc)^2+(ab)^2}{ (abc)^2 }\] we need to get this with all the info

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)\]

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so the answer to\[[3(\frac{ (ab)^2(bc)^2+(ca))^2 }{ (abc)^2 })]3=[3\frac{ 2 }{ 1 }]3=2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wat doyou think guys,please help

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how can we find the answer 1

anonymous
 4 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i am out of medals...thanks

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Multiply 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i really went for miles to show a centimeter journey

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Now and again, it is just a question of doing the first thing that comes to mind....:)

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0We are not finding solutions, we are evaluating the expression in the top post...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is it possible to getthe values ofa,b,c

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0very helpful,i asked this becuause 1 was looking for the solution of the cube

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://en.wikipedia.org/wiki/Cubic_function#Reduction_to_a_depressed_cubic
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