What is the number of solution of the equation $\ x^3 - \lfloor x\rfloor = 3$? What is the number of solution of the equation $\ x^3 - \lfloor x\rfloor = 3$ ? (where $\lfloor x \rfloor\ $ is the greatest integer $\le x$)
I tried plotting the graphs of these equations on Desmos graph calculator and that they intersect each other in between x = 1 and 2 but I couldn't figure out a way to get to this conclusion on my own.
Is there any way by which I can determine where these functions intersect?

First of all $x > 0$.
If $x \le 0$ then $x^3 < 0$ so $[x] = x^3 -3 \le -3$ so $[x] \le -3$ and $x \le -2$.  (Obviously I'm being overly cautious. $[x]\ne x$ unless $x$ is on integer so $x \le -3$ and obviously $x$ can't be $0$ so $[x]\le -4$ and ... so on but... this just stick with $x \le -2$.)
This means that $|x^3| > |x|$ and that $x^3 - x < 0$.  So $x^3 - [x] \ge x^3 -x > x^3 -[x] -1$ so $x^3 -[x] < 1< 3$.
So $x > 0$.
....
Now $[x] \le x$ so $[x]^3 - [x] \le x^3-[x] = 3$.
We must have $[x]^3 -[x] = [x]([x]^2 - 1) = [x]([x]-1)([x]+1) \le 3$.
If $[x] \ge 2$ then $[x]([x]-1)([x]+1)\ge 2\cdot 1\cdot 3 > 3$ so we must have
$[x]=1$
Okay.  So $1\le x < 2$.  Let $r = x-1$ and $x = 1+ r$.
Then $x^3 -[x] = (1+r)^3 =1 = 3r + 3r^2 + r^3 = 3$.
Now $r$ is monotonically increasing on $r\in [0,1)$ so there is at most one solution.
And $3\cdot 0 + 3\cdot 0^2 + 0^3 < 3$ and $\lim_{r\to 1} 3r + 3r^2 + r^3 = 7 > 3$.
So by continuity there is one solution.
We might be able to solve for $r$ but we don't have to.  $r$ is the unique and only solution to $3r + 3r^2 + r^3 = 3$ and $x = 1+r$.
And... oh, googly moogly......
If $[x]=1$ then $x^3 - 1 =3$ and..... $x =\sqrt[3] 4$ and $r = \sqrt[3] 4 -1$  is the unique solution to $3r+3r^2 +r^3 = 3$ and ....
Ah... geeze.... that was trivial to solvi
$3r + 3r^2 + r^3 = 3$
$1+3r + 3r^2 + r^3 = 3+1 = 4$.
$(1+r)^3 = 4$.
.... forest for the trees, forest for the trees.
