<s>[INST] Finding sum of a series: difference of cubes I am trying to find sum of the infinite series:  
$$1^{3}-2^{3}+3^{3}-4^{3}+5^{3}-6^{3} + \ldots$$
I tried to solve it by subtracting sum of even cubes from odd, but that solves only half of the numbers.   
Any input is appreciated.
Thank you all (especially Marvis, Manzonni, Gottfried, Mhenni for long and descriptive answers) for you answer. It's my first visit to math.stackexchange.com. Very impressed with all the help. :)
Now, I would like to know how Marvis got to partial sum formula:
$S_n = \sum_{k=0}^{n}(-1)^{k+1} k^3 = \dfrac{1 + (-1)^n(4n^3 + 6n^2-1)}8 = \begin{cases} \dfrac{2n^3+3n^2}4; & n \text{ is even}\\ \dfrac{1-3n^2-2n^3}4; & n \text{ is odd} \end{cases}$
I tried to replicate it, but can't get to it.  
 [/INST] Using the identities
$$
\sum_{k=0}^\infty\binom{k}{n}x^k=\frac{x^n}{(1-x)^{n+1}}
$$
and
$$
k^3=\binom{k}{1}+6\binom{k}{2}+6\binom{k}{3}
$$
yields
$$
\sum_{k=1}^\infty k^3x^{k-1}=\frac{1}{(1-x)^2}+6\frac{x}{(1-x)^3}+6\frac{x^2}{(1-x)^4}
$$
Letting $x\to-1$, gives
$$
\begin{align}
\sum_{k=0}^\infty(-1)^{k-1}k^3
&=\frac14-6\cdot\frac18+6\cdot\frac1{16}\\
&=-\frac18
\end{align}
$$
</s>