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*}