Question

x^3-x-1=0,find the value of

Answer 1

\[\frac{ 1-a }{ 1+a }+\frac{ 1-b }{ 1+b}+\frac{ 1-c }{ 1+c }\]
if \[a,b,c \] are roots of the equation

Answer 2

\[\frac{ 1 }{ 1+a }+\frac{ 1 }{ 1+b }+\frac{ 1 }{ 1+c }-3\]
if we sibstitute all roots individually\[a^3=a+1,b^3=b+1,c^3=c+1\]
hence
\[\frac{ 1}{ a^3 }+\frac{ 1 }{ b^3 }+\frac{ 1 }{ c^3 }-3\]