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## Jonask Group Title if one year ago one year ago

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1. estudier Group Title

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 Group Title

$a,b,c$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 Group Title

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 Group Title

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

5. Jonask Group Title

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

6. Jonask Group Title

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

7. Jonask Group Title

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

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

9. Jonask Group Title

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

10. Jonask Group Title

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

11. Jonask Group Title

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

12. Jonask Group Title

wat doyou think guys,please help

13. Zekarias Group Title

Actually the answer is 1 So you are not correct

14. Jonask Group Title

how can we find the answer 1

15. Zekarias Group Title

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

16. Jonask Group Title

i am out of medals...thanks

17. estudier Group Title

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 Group Title

i really went for miles to show a centimeter journey

19. estudier Group Title

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

20. UnkleRhaukus Group Title

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

21. estudier Group Title

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

22. UnkleRhaukus Group Title

to hard for me

23. Jonask Group Title

is it possible to getthe values ofa,b,c

24. estudier Group Title

Do you mean by hand?

25. estudier Group Title
26. Jonask Group Title

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

27. estudier Group Title