<s>[INST] Find the nth term of the given series The series is $$1^3 + (2^3 +3^3)+(4^3+5^3+6^3)+...$$
i.e., $1^3$, $(2^3 +3^3)$ , $(4^3+5^3+6^3)$ , $(7^3+8^3+9^3+10^3)$ , and so on are the $1^{st}$, $2^{nd}$, $3^{rd}$ and so on terms respectively. ($(4^3+5^3+6^3)$ is the third term). So we need to find the $n^{th}$ term.
So, the first element of every individual term makes a sequence $1, 2, 4, 7...$.  $\therefore$ the first element of each term is $(1+\frac{n(n-1)}2)^3$ and this the last element is $(\frac{n(n-1)}2 + 1 + (n-1))^3$. But I don't understand what to do next.
 [/INST] Let me see if I understand you correctly for I am a little bit confused about your notation. I guess that by the first term of the sequence you mean $1^3$ and $(2^3 + 3^3)$ is the second term, $(4^3 + 5^3 + 6^3)$ is the third term and so on. If this is so, it seems correct that the firs summand of the $n$-th term is given by 
$$(\frac{n(n-1)}{2}+1)^3 \ .$$ 
Since the $n$-th term contains $n$ summands, the last summand of the $n$-th term is given by 
$$(\frac{n(n-1)}{2}+1 + (n-1))^3 = (\frac{n(n+1)}{2})^3 \ , $$
where we added $(n-1)$ because the first summand is to be raised $(n-1)$ times by $1$ to get the last summand of the $n$-th term (your formula is almost the same for it seems that you used the same logic and only forgot to add $+1$). This expression will serve as a checker after we express the general formula for the $n$-th term.
To describe the $n$-th term should be straightforward now for we know the first summand and the number of steps we have to make to get to the last summand for arbitrary $n \in \mathbb{N}$. The $n$-th term should be 
$$ \sum_{i=0}^{n-1} (\frac{n(n-1)}{2}+1 +i)^3 \ . $$ 
The first summand is obviously in accordance with what we know and if we compute the last summand of the above expression we get
$$ (\frac{n(n-1)}{2}+1 + n-1)^3 = (\frac{n(n-1)}{2} + n)^3 = (\frac{n(n+1)}{2} )^3 \ ,$$ 
which is what we expected. 
In case this answer is not what you were looking for, try to specify more carefully what you mean by $n$-th term.
</s>