Answer:
See explaination
Explanation:
given f:R-\left \{ 1 \right \}\rightarrow R-\left \{ 1 \right \} defined by f(x)=\left ( \frac{x+1}{x-1} \right )^{3}
let f(x)=f(y)
\left ( \frac{x+1}{x-1} \right )^{3}=\left ( \frac{y+1}{y-1} \right )^{3}
taking cube roots on both sides , we get
\frac{x+1}{x-1} = \frac{y+1}{y-1}
\Rightarrow (x+1)(y-1)=(x-1)(y+1)
\Rightarrow xy-x+y-1=xy+x-y-1
\Rightarrow -x+y=x-y
\Rightarrow x+x=y+y
\Rightarrow 2x=2y
\Rightarrow x=y
Hence f is one - one
let y\in R, such that f(x)=\left ( \frac{x+1}{x-1} \right )^{3}=y
\Rightarrow \frac{x+1}{x-1} =\sqrt[3]{y}
\Rightarrow x+1=\sqrt[3]{y}\left ( x-1 \right )
\Rightarrow x+1=\sqrt[3]{y} x- \sqrt[3]{y}
\Rightarrow \sqrt[3]{y} x-x=1+ \sqrt[3]{y}
\Rightarrow x\left ( \sqrt[3]{y} -1 \right ) =1+ \sqrt[3]{y}
\Rightarrow x=\frac{\sqrt[3]{y}+1}{\sqrt[3]{y}-1}
for every y\in R-\left \{ 1 \right \}\exists x\in R-\left \{ 1 \right \} such that x=\frac{\sqrt[3]{y}+1}{\sqrt[3]{y}-1}
Hence f is onto
since f is both one -one and onto so it is a bijective