Question

if

Answer 1

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 ...

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

\[(x-a)(x-b)(x-c)=x^3-x^2(a+b+c)+x(bc+ab+ac)-(abc)\]
coefeecients of x^2 is 0
x is -1 and constant 0 so\[a+b+c=0,bc+ab+ac=-1,abc=1\]

Answer 8

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

Answer 9

\[(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)\]

Answer 10

\[(0)^2=(ab)^2+(bc)^2+(ca)^2+2(-1)\]
\[(ab)^2+(bc)^2+(ca)^2=2\]

Answer 11

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

Answer 12

wat doyou think guys,please help

Answer 13

Actually the answer is 1
So you are not correct

Answer 14

how can we find the answer 1

Answer 15

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

Answer 16

i am out of medals...thanks

Answer 17

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

Answer 18

i really went for miles to show a centimeter journey

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

to hard for me

Answer 23

is it possible to getthe values ofa,b,c

Answer 24

Do you mean by hand?

Answer 25

http://en.wikipedia.org/wiki/Plastic_number

Answer 26

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

Answer 27

http://en.wikipedia.org/wiki/Cubic_function#Reduction_to_a_depressed_cubic