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Jonask

  • 2 years ago

if

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  1. estudier
    • 2 years ago
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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 ...

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

  3. Jonask
    • 2 years ago
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    simplified \[\frac{ 1 }{ a+1 }+\frac{ 1 }{ b+1 }+\frac{ 1 }{ c+1 }-3\] note\[a^3=a+1,b^3=b+1,c^3=c+1\]

  4. Jonask
    • 2 years ago
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    so\[\frac{ 1 }{ a^3 }+\frac{ 1 }{ b^3 }+\frac{ 1 }{ c^3 }-3\]

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

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

  7. Jonask
    • 2 years ago
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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\[a+b+c=0,bc+ab+ac=-1,abc=1\]

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

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

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

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

  12. Jonask
    • 2 years ago
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    wat doyou think guys,please help

  13. Zekarias
    • 2 years ago
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    Actually the answer is 1 So you are not correct

  14. Jonask
    • 2 years ago
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    how can we find the answer 1

  15. Zekarias
    • 2 years ago
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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

  16. Jonask
    • 2 years ago
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    i am out of medals...thanks

  17. estudier
    • 2 years ago
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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

  18. Jonask
    • 2 years ago
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    i really went for miles to show a centimeter journey

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

  20. UnkleRhaukus
    • 2 years ago
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    \[x^3−x−1=0\] 1 is not a solution as \[1^3−1−1=1\neq0\]

  21. estudier
    • 2 years ago
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    We are not finding solutions, we are evaluating the expression in the top post...

  22. UnkleRhaukus
    • 2 years ago
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    to hard for me

  23. Jonask
    • 2 years ago
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    is it possible to getthe values ofa,b,c

  24. estudier
    • 2 years ago
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    Do you mean by hand?

  25. estudier
    • 2 years ago
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    http://en.wikipedia.org/wiki/Plastic_number

  26. Jonask
    • 2 years ago
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    very helpful,i asked this becuause 1 was looking for the solution of the cube

  27. estudier
    • 2 years ago
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    http://en.wikipedia.org/wiki/Cubic_function#Reduction_to_a_depressed_cubic

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