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
 A: Since $$ x^3-x-6< x^3-x-5\leq 0$$ we have $$(x-2)(x^2+2x+3)<0$$
Since $x^2+2x+3 = (x+1)^2+2 $ we have $ \boxed{x<2}$.
On the other side we have $$0<x^3-x-4<x(x^2-1)$$
so $\boxed{x\in (-1,0)\cup (1,\infty)}$. Both together we get $$\boxed{x\in (-1,0)\cup (1,2)}$$
Now if $x\in (-1,0)$ then $\lfloor x\rfloor = -1$ so $x^3 = 4$ so no solution. 
and if $x\in (1,2)$ then $\lfloor x\rfloor=1$ so $x^3= 6$ so $x=\sqrt[3]{6}$ is a solution.
A: 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 $x=\sqrt[3]{6}$

A: Consider the three functions $ y = x-1 $, $ y = x ^ 3-5 $ and $ y = x $. Compare the portion of the graph of $ y = x ^ 3-5 $ that is between the graphs of $ y = x-1 $ and $ y = x $. What can you say about that?
The possible solutions of the equation $x^3-\lfloor x\rfloor=5 $ live in the set of solutions of the inequation $x-1<x^3-5\leq x$.
From the graph we can see that all solutions of the equation $x^3-\lfloor x\rfloor=5 $ are larger than the only real root of $x^3-5= x-1$ and less than the only real root of $x^3-5=x$.
