anonymous
  • anonymous
if
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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 ...
anonymous
  • anonymous
\[a,b,c \]are roots of the equation\[x^3-x-1=0\] compute \[\frac{ 1-a }{ 1+a }+\frac{ 1-b }{ 1+b }+\frac{ 1-c }{ 1+c}\]
anonymous
  • anonymous
simplified \[\frac{ 1 }{ a+1 }+\frac{ 1 }{ b+1 }+\frac{ 1 }{ c+1 }-3\] note\[a^3=a+1,b^3=b+1,c^3=c+1\]

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anonymous
  • anonymous
so\[\frac{ 1 }{ a^3 }+\frac{ 1 }{ b^3 }+\frac{ 1 }{ c^3 }-3\]
anonymous
  • anonymous
if we divide by a,b,c\[1-\frac{ 1 }{ a^2 }=\frac{ 1 }{ a^3 }\]
anonymous
  • anonymous
\[[3-(\frac{ 1 }{ a^2 }+\frac{ 1 }{ b^2 }+\frac{ 1 }{ c^2 })]-3\]
anonymous
  • anonymous
\[(x-a)(x-b)(x-c)=x^3-x^2(a+b+c)+x(bc+ab+ac)-(abc)\] coefeecients of x^2 is 0 x is -1 and constant 0 so\[a+b+c=0,bc+ab+ac=-1,abc=1\]
anonymous
  • anonymous
\[\frac{ (ac)^2+(bc)^2+(ab)^2}{ (abc)^2 }\] we need to get this with all the info
anonymous
  • anonymous
\[(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)\]
anonymous
  • anonymous
\[(0)^2=(ab)^2+(bc)^2+(ca)^2+2(-1)\] \[(ab)^2+(bc)^2+(ca)^2=2\]
anonymous
  • anonymous
so the answer to\[[3-(\frac{ (ab)^2(bc)^2+(ca))^2 }{ (abc)^2 })]-3=[3-\frac{ 2 }{ 1 }]-3=-2\]
anonymous
  • anonymous
wat doyou think guys,please help
anonymous
  • anonymous
Actually the answer is 1 So you are not correct
anonymous
  • anonymous
how can we find the answer 1
anonymous
  • anonymous
\[\frac{ 1-a }{ 1+a }+\frac{ 1-b }{ 1+b }+\frac{ 1-c }{ 1+c }\] \[-1+\frac{ 2 }{ 1+a }-1+\frac{ 2 }{ 1+b }-1+\frac{ 2 }{ 1+c }\] \[2(\frac{ 1 }{ 1+a }+\frac{ 1 }{ 1+b }+\frac{ 1 }{ 1+c })-3\] \[2(\frac{ (1+b)(1+c)+(1+a)(1+c)+(1+a)(1+b) }{ (1+a)(1+b)(1+c) })-3\] \[2(\frac{ (ab+ac+bc)+2(a+b+c)+3 }{ abc+(ab+ac+bc)+(a+b+c)+1 })-3\]Now insert the respective values so as to get the ANSWER 1
anonymous
  • anonymous
i am out of medals...thanks
anonymous
  • anonymous
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
anonymous
  • anonymous
i really went for miles to show a centimeter journey
anonymous
  • anonymous
Now and again, it is just a question of doing the first thing that comes to mind....:-)
UnkleRhaukus
  • UnkleRhaukus
\[x^3−x−1=0\] 1 is not a solution as \[1^3−1−1=1\neq0\]
anonymous
  • anonymous
We are not finding solutions, we are evaluating the expression in the top post...
UnkleRhaukus
  • UnkleRhaukus
to hard for me
anonymous
  • anonymous
is it possible to getthe values ofa,b,c
anonymous
  • anonymous
Do you mean by hand?
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Plastic_number
anonymous
  • anonymous
very helpful,i asked this becuause 1 was looking for the solution of the cube
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Cubic_function#Reduction_to_a_depressed_cubic

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