<s>[INST] Expressing $\sin^4x-\sin^6x$ in another way I am slightly confused on how one would subtract $\sin^4x-\sin^6x$.
I know that $\sin^2x=(1/2)(1-\cos2x)$,
so $\sin^4x$ would logically be $[(1/2)(1-\cos2x)]^2=(1/4)(1-2\cos(2x)+\cos^2(2x)$
However the value of $\sin^6x$ eludes me. Would it be $(1-\cos2x)^3$? I did that and got $1-\cos2x-2\cos2x-2\cos^2(2x)+\cos^2(2x)-\cos^3(2x)$
I am not sure if that is correct. 
 [/INST] I have noticed the other answers either don't answer your question (and get many upvotes none-the-less) or are much more complicated than necessary (even more complicated than you understand based on the comment by Raymond).  So, here is the answer to your actual question.
You dropped the 1/2.
$$\sin^6 x = (\sin^2 x)^3 = \left(\frac{1 - \cos{2x}}{2}\right)^3 = \frac{1}{8} (1 - \cos{2x})^3$$
so you end up missing $1/8$ in the end.  Also, you messed up the sign on one term.  In general
$$(A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3$$
so here we have
\begin{align*}
  (1 - \cos(2x))^3 &= \bigg(1 + (-\cos(2x))\bigg)^3 \\
  &= 1 + 3(-\cos(2x)) + 3(-\cos(2x))^2 + (-\cos(2x))^3 \\
  &= 1 - 3\cos(2x) + 3\cos^2(2x) - \cos^3(2x).
\end{align*}
Therefore, your final answer is
\begin{align*}
  \sin^4x - \sin^6x &= \frac{1}{4}\bigg(1 - 2\cos(2x) + \cos^2(2x)\bigg) \\
  &- \frac{1}{8}\bigg(1 - 3\cos(2x) + 3\cos^2(2x) - \cos^3(2x)\bigg) \\
  &= \frac{1}{8}\bigg(1  - \cos(2x) - \cos^2(2x) + \cos^3(2x)\bigg)
\end{align*}
</s>