For any real number $a$, let $\lfloor a\rfloor$ denote the greatest integer not exceeding $a$. In positive real numbers solve the following equation

$$
n+\lfloor\sqrt{n}\rfloor+\lfloor\sqrt[3]{n}\rfloor=2014
$$
Obviously $n$ must be a positive integer. Now note that $44^{2}=1936<2000$, than $2014=n+\lfloor\sqrt{n}\rfloor+\lfloor\sqrt[3]{n}\rfloor>2000+44+12=2056$, a contradiction!

So $1950 \leq n \leq 2000$, therefore $\lfloor\sqrt{n}\rfloor=44$ and $\lfloor\sqrt[3]{n}\rfloor=12$. Plugging that into the original equation we get:

$$
n+\lfloor\sqrt{n}\rfloor+\lfloor\sqrt[3]{n}\rfloor=n+44+12=2014
$$

From which we get $n=1958$, which is the only solution.

Solution2. Obviously $n$ must be a positive integer. Since $n \leq 2014, \sqrt{n} \leq \sqrt{2014} < 45$ and $\sqrt[3]{n} \leq \sqrt[3]{2014} < 13$, thus $\lfloor\sqrt{n}\rfloor=44$ and $\lfloor\sqrt[3]{n}\rfloor=12$ and $n=2014-\lfloor\sqrt{n}\rfloor-\lfloor\sqrt[3]{n}\rfloor=2014-44-12=1958$.