Let the real number $x$ satisfies $\left\{ x \right\} + \left\{ {\frac{1}{x}} \right\} = 1$ and $k$ denotes the value of $\left\{ {{x^3}} \right\} + \left\{ {\frac{1}{{{x^3}}}} \right\} = k$ (where $\{\cdot\}$ denotes fractional part function). Find the value of $k$.
My approach is as follows: $x$ cannot be an integer, $x = n + a,a \in \left( 0,1 \right)$
$$a + \left\{ {\frac{1}{{n + a}}} \right\} = 1 \Rightarrow \left\{ {\frac{1}{{n + a}}} \right\} = 1 - a = t,t \in \left( {0,1} \right)$$
Not able to proceed from here.
Hint: Since $\{x\}+\{x^{-1}\}$ is an integer, that means $x+x^{-1}$ is also an integer. So $x^3+x^{-3}=(x+x^{-1})^3-3(x+x^{-1})$ is also an integer. Now show $x^3$ is irrational to conclude $\{x^3\}+\{x^{-3}\}=1$.