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Jonask Group Title

if

  • one year ago
  • one year ago

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  1. estudier Group Title
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    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
  2. Jonask Group Title
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    \[a,b,c \]are roots of the equation\[x^3-x-1=0\] compute \[\frac{ 1-a }{ 1+a }+\frac{ 1-b }{ 1+b }+\frac{ 1-c }{ 1+c}\]

    • one year ago
  3. Jonask Group Title
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    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
  4. Jonask Group Title
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    so\[\frac{ 1 }{ a^3 }+\frac{ 1 }{ b^3 }+\frac{ 1 }{ c^3 }-3\]

    • one year ago
  5. Jonask Group Title
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    if we divide by a,b,c\[1-\frac{ 1 }{ a^2 }=\frac{ 1 }{ a^3 }\]

    • one year ago
  6. Jonask Group Title
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    \[[3-(\frac{ 1 }{ a^2 }+\frac{ 1 }{ b^2 }+\frac{ 1 }{ c^2 })]-3\]

    • one year ago
  7. Jonask Group Title
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    \[(x-a)(x-b)(x-c)=x^3-x^2(a+b+c)+x(bc+ab+ac)-(abc)\] coefeecients of x^2 is 0 x is -1 and constant 0 so\[a+b+c=0,bc+ab+ac=-1,abc=1\]

    • one year ago
  8. Jonask Group Title
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    \[\frac{ (ac)^2+(bc)^2+(ab)^2}{ (abc)^2 }\] we need to get this with all the info

    • one year ago
  9. Jonask Group Title
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    \[(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)\]

    • one year ago
  10. Jonask Group Title
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    \[(0)^2=(ab)^2+(bc)^2+(ca)^2+2(-1)\] \[(ab)^2+(bc)^2+(ca)^2=2\]

    • one year ago
  11. Jonask Group Title
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    so the answer to\[[3-(\frac{ (ab)^2(bc)^2+(ca))^2 }{ (abc)^2 })]-3=[3-\frac{ 2 }{ 1 }]-3=-2\]

    • one year ago
  12. Jonask Group Title
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    wat doyou think guys,please help

    • one year ago
  13. Zekarias Group Title
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    Actually the answer is 1 So you are not correct

    • one year ago
  14. Jonask Group Title
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    how can we find the answer 1

    • one year ago
  15. Zekarias Group Title
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    \[\frac{ 1-a }{ 1+a }+\frac{ 1-b }{ 1+b }+\frac{ 1-c }{ 1+c }\] \[-1+\frac{ 2 }{ 1+a }-1+\frac{ 2 }{ 1+b }-1+\frac{ 2 }{ 1+c }\] \[2(\frac{ 1 }{ 1+a }+\frac{ 1 }{ 1+b }+\frac{ 1 }{ 1+c })-3\] \[2(\frac{ (1+b)(1+c)+(1+a)(1+c)+(1+a)(1+b) }{ (1+a)(1+b)(1+c) })-3\] \[2(\frac{ (ab+ac+bc)+2(a+b+c)+3 }{ abc+(ab+ac+bc)+(a+b+c)+1 })-3\]Now insert the respective values so as to get the ANSWER 1

    • one year ago
  16. Jonask Group Title
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    i am out of medals...thanks

    • one year ago
  17. estudier Group Title
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    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
  18. Jonask Group Title
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    i really went for miles to show a centimeter journey

    • one year ago
  19. estudier Group Title
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    Now and again, it is just a question of doing the first thing that comes to mind....:-)

    • one year ago
  20. UnkleRhaukus Group Title
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    \[x^3−x−1=0\] 1 is not a solution as \[1^3−1−1=1\neq0\]

    • one year ago
  21. estudier Group Title
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    We are not finding solutions, we are evaluating the expression in the top post...

    • one year ago
  22. UnkleRhaukus Group Title
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    to hard for me

    • one year ago
  23. Jonask Group Title
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    is it possible to getthe values ofa,b,c

    • one year ago
  24. estudier Group Title
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    Do you mean by hand?

    • one year ago
  25. estudier Group Title
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    http://en.wikipedia.org/wiki/Plastic_number

    • one year ago
  26. Jonask Group Title
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    very helpful,i asked this becuause 1 was looking for the solution of the cube

    • one year ago
  27. estudier Group Title
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    http://en.wikipedia.org/wiki/Cubic_function#Reduction_to_a_depressed_cubic

    • one year ago
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