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## Jonask 2 years ago if

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1. estudier

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

$a,b,c$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

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

so$\frac{ 1 }{ a^3 }+\frac{ 1 }{ b^3 }+\frac{ 1 }{ c^3 }-3$

5. Jonask

if we divide by a,b,c$1-\frac{ 1 }{ a^2 }=\frac{ 1 }{ a^3 }$

6. Jonask

$[3-(\frac{ 1 }{ a^2 }+\frac{ 1 }{ b^2 }+\frac{ 1 }{ c^2 })]-3$

7. Jonask

$(x-a)(x-b)(x-c)=x^3-x^2(a+b+c)+x(bc+ab+ac)-(abc)$ 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

$\frac{ (ac)^2+(bc)^2+(ab)^2}{ (abc)^2 }$ we need to get this with all the info

9. Jonask

$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$

10. Jonask

$(0)^2=(ab)^2+(bc)^2+(ca)^2+2(-1)$ $(ab)^2+(bc)^2+(ca)^2=2$

11. Jonask

so the answer to$[3-(\frac{ (ab)^2(bc)^2+(ca))^2 }{ (abc)^2 })]-3=[3-\frac{ 2 }{ 1 }]-3=-2$

12. Jonask

wat doyou think guys,please help

13. Zekarias

Actually the answer is 1 So you are not correct

14. Jonask

how can we find the answer 1

15. Zekarias

$\frac{ 1-a }{ 1+a }+\frac{ 1-b }{ 1+b }+\frac{ 1-c }{ 1+c }$ $-1+\frac{ 2 }{ 1+a }-1+\frac{ 2 }{ 1+b }-1+\frac{ 2 }{ 1+c }$ $2(\frac{ 1 }{ 1+a }+\frac{ 1 }{ 1+b }+\frac{ 1 }{ 1+c })-3$ $2(\frac{ (1+b)(1+c)+(1+a)(1+c)+(1+a)(1+b) }{ (1+a)(1+b)(1+c) })-3$ $2(\frac{ (ab+ac+bc)+2(a+b+c)+3 }{ abc+(ab+ac+bc)+(a+b+c)+1 })-3$Now insert the respective values so as to get the ANSWER 1

16. Jonask

i am out of medals...thanks

17. estudier

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

i really went for miles to show a centimeter journey

19. estudier

Now and again, it is just a question of doing the first thing that comes to mind....:-)

20. UnkleRhaukus

$x^3−x−1=0$ 1 is not a solution as $1^3−1−1=1\neq0$

21. estudier

We are not finding solutions, we are evaluating the expression in the top post...

22. UnkleRhaukus

to hard for me

23. Jonask

is it possible to getthe values ofa,b,c

24. estudier

Do you mean by hand?

25. estudier
26. Jonask

very helpful,i asked this becuause 1 was looking for the solution of the cube

27. estudier

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