anonymous 4 years ago if

1. 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 ...

2. anonymous

$a,b,c$are roots of the equation$x^3-x-1=0$ compute $\frac{ 1-a }{ 1+a }+\frac{ 1-b }{ 1+b }+\frac{ 1-c }{ 1+c}$

3. anonymous

simplified $\frac{ 1 }{ a+1 }+\frac{ 1 }{ b+1 }+\frac{ 1 }{ c+1 }-3$ note$a^3=a+1,b^3=b+1,c^3=c+1$

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. 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$a+b+c=0,bc+ab+ac=-1,abc=1$

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. anonymous

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

12. anonymous

13. anonymous

Actually the answer is 1 So you are not correct

14. anonymous

how can we find the answer 1

15. 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

16. anonymous

i am out of medals...thanks

17. 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

18. anonymous

i really went for miles to show a centimeter journey

19. anonymous

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. anonymous

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

22. UnkleRhaukus

to hard for me

23. anonymous

is it possible to getthe values ofa,b,c

24. anonymous

Do you mean by hand?

25. anonymous
26. anonymous

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

27. anonymous