Can both $n+3\; \text{and}\; n^2+3$ both be cubic numbers at same time? 
Can both $n+3\; \text{and}\; n^2+3$ both be cubic number at same time? Where $n$ is an integer number. Not necessarily positive.    

I tried writing $x^3 = n+3$ and expressing $n^2+3$ in terms of $x$. I found $x^6 -6x^3+12$ but this doesn't help. How do I prove this?

Your question is asking if there is $x,a \in \mathbb{Z}$ such that $$x^6-6x^3+12=a^3$$
Note that if we have $x \ge 2, x \le -2$, then we have that $$(x^2-1)^3 = x^6-3x^4+3x^2-1 < x^6-6x^3+12=a^3$$
And also $$(x^2+1)^3 =x^6+3x^4+3x^2+1 > x^6-6x^3+12=a^3$$
This give us $x^2-1<a<x^2+1$, thus forcing $a$ to be $x^2$. However, as $x^3 \neq 2$ for any integer $x$, this is a contradiction. 
The only cases left are $x=-1, 0,1 $, which can be manually checked to be never be cubes. 
