Here's the question you clicked on:
Jonask
if
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 ...
\[a,b,c \]are roots of the equation\[x^3-x-1=0\] compute \[\frac{ 1-a }{ 1+a }+\frac{ 1-b }{ 1+b }+\frac{ 1-c }{ 1+c}\]
simplified \[\frac{ 1 }{ a+1 }+\frac{ 1 }{ b+1 }+\frac{ 1 }{ c+1 }-3\] note\[a^3=a+1,b^3=b+1,c^3=c+1\]
so\[\frac{ 1 }{ a^3 }+\frac{ 1 }{ b^3 }+\frac{ 1 }{ c^3 }-3\]
if we divide by a,b,c\[1-\frac{ 1 }{ a^2 }=\frac{ 1 }{ a^3 }\]
\[[3-(\frac{ 1 }{ a^2 }+\frac{ 1 }{ b^2 }+\frac{ 1 }{ c^2 })]-3\]
\[(x-a)(x-b)(x-c)=x^3-x^2(a+b+c)+x(bc+ab+ac)-(abc)\] coefeecients of x^2 is 0 x is -1 and constant 0 so\[a+b+c=0,bc+ab+ac=-1,abc=1\]
\[\frac{ (ac)^2+(bc)^2+(ab)^2}{ (abc)^2 }\] we need to get this with all the info
\[(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)\]
\[(0)^2=(ab)^2+(bc)^2+(ca)^2+2(-1)\] \[(ab)^2+(bc)^2+(ca)^2=2\]
so the answer to\[[3-(\frac{ (ab)^2(bc)^2+(ca))^2 }{ (abc)^2 })]-3=[3-\frac{ 2 }{ 1 }]-3=-2\]
wat doyou think guys,please help
Actually the answer is 1 So you are not correct
how can we find the answer 1
\[\frac{ 1-a }{ 1+a }+\frac{ 1-b }{ 1+b }+\frac{ 1-c }{ 1+c }\] \[-1+\frac{ 2 }{ 1+a }-1+\frac{ 2 }{ 1+b }-1+\frac{ 2 }{ 1+c }\] \[2(\frac{ 1 }{ 1+a }+\frac{ 1 }{ 1+b }+\frac{ 1 }{ 1+c })-3\] \[2(\frac{ (1+b)(1+c)+(1+a)(1+c)+(1+a)(1+b) }{ (1+a)(1+b)(1+c) })-3\] \[2(\frac{ (ab+ac+bc)+2(a+b+c)+3 }{ abc+(ab+ac+bc)+(a+b+c)+1 })-3\]Now insert the respective values so as to get the ANSWER 1
i am out of medals...thanks
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
i really went for miles to show a centimeter journey
Now and again, it is just a question of doing the first thing that comes to mind....:-)
\[x^3−x−1=0\] 1 is not a solution as \[1^3−1−1=1\neq0\]
We are not finding solutions, we are evaluating the expression in the top post...
is it possible to getthe values ofa,b,c
very helpful,i asked this becuause 1 was looking for the solution of the cube
http://en.wikipedia.org/wiki/Cubic_function#Reduction_to_a_depressed_cubic