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

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.
