Solve for x: $x^3-\lfloor x\rfloor=5$ 
Solve for x:$$x^3-\lfloor x\rfloor=5$$

My Attempt:
$x^3-5=\lfloor x\rfloor$
Now, $x-1<\lfloor x\rfloor\leq x$
$x-1<x^3-5\leq x$
Not able to proceed from here

Write $\lfloor{x}\rfloor$ as $x-\epsilon$ where $\epsilon$ is the fractional part, so $0\le\epsilon<1$. Now consider the function $f(x)=x^3-x-5+\epsilon$, note that $f$ has at least one root since $f$ is cubic polynomial. Now differentiate and get $f’(x)=3x^2-1$ which has two roots $\pm\frac{1}{\sqrt{3}}$.
Now:

$f(\frac{-1}{\sqrt{3}})=-\frac13+\frac{1}{\sqrt{3}}-5+a<0$ thus no solution for $x\le \frac{-1}{\sqrt3}$
$f(\frac{1}{\sqrt{3}})=\frac13-\frac{1}{\sqrt{3}}-5+a<0$ thus no solution for $-\frac{1}{\sqrt3}\le x \frac{1}{\sqrt3}$
Finally $f$ has one single root. Since $f(1)<0$ and $f(2)>0$ the root is located in $(1,2)$ thus $\lfloor{x}\rfloor=1$ wich gives $x=\sqrt[3]{6}$

