Using generating functions find the sum $1^3 + 2^3 + 3^3 +\dotsb+ n^3$ I am quite new to generating functions concept and I am really finding it difficult to know how to approach problems like this. I need to find the sum of $1^3 + 2^3 + 3^3 +\dotsb+ n^3$ using generating functions. How do I proceed about it?

If you really want to get blown away, consider the following, taken from Aigner's "A Course in Enumeration" (Springer, 2007). Define:
$$
s_m(n) = \sum_{0 \le k < n} k^m
$$
and it's exponential generating function:
$
  \begin{align}
    \widehat{S}_n(z)
      &= \sum_{m \ge 0} s_m(n) \frac{z^m}{m!} \\
      &= \sum_{1 \le k \le n - 1} \sum_{m \ge 0} \frac{k^m z^m}{m!} \\
      &= \sum_{1 \le k \le n - 1} \mathrm{e}^{k z} \\
      &= \frac{\mathrm{e}^{n z} - 1}{\mathrm{e}^z - 1}
  \end{align}
$
This is almost the exponential generating function for the powers:
$
  \begin{align}
    \widehat{P}(z)
      &= \sum_{m \ge 0} n^m \frac{z^m}{m!} \\
      &= \mathrm{e}^{n z}
  \end{align}
$
Sadly, the series $\mathrm{e}^z - 1$ has no reciprocal, as it has no constant term. But we can write:
$$
    (\widehat{P}(z) - 1) \widehat{B}(z)
    = z \widehat{S}(z)
$$
where:
$$
    \widehat{B}(z)
      = \frac{z}{\mathrm{e}^z - 1}
$$
whose coefficients are the Bernoulli numbers:
$\begin{align}
    \widehat{B}(z)
      &= \sum_{n \ge 0} B_n \frac{z^n}{n!} \\
      &= 1 - \frac{1}{2} z + \frac{1}{6} \frac{z^2}{2!}
    - \frac{1}{30} \frac{z^4}{4!} + \frac{1}{42} \frac{z^6}{6!}
    - \frac{1}{30} \frac{z^8}{8!} + \frac{5}{66} \frac{z^{10}}{10!}
    - \frac{691}{2130} \frac{z^{12}}{12!}
    + \frac{7}{6} \frac{z^{14}}{14!}
    - \dotsb
\end{align}$
and finally:
$$
    \sum_{m \ge 0} s_m(n) \frac{z^{m + 1}}{m!}
      = \sum_{m \ge 0} z^m
   \sum_{0 \le k \le m}
     \binom{m}{k} \frac{(n z)^{m - k}}{(m - k)!} B_k
$$
Comparing coefficients of $z^{m + 1}$ and simplifying:
$$
    s_m(n)
      = \frac{1}{m + 1}
    \sum_{0 \le k \le m} (-1)^k \binom{m + 1}{k} B_k n^{m + 1 - k}
$$
