Prove that $\sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}}=\frac{1}{4}(1+3\cos^2\theta)$ A question on submultiple angles states...

Prove that:$$\sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}}=\frac{1}{4}(1+3\cos^2\theta)$$

My efforts
I tried using the formula
$$a^3+b^3=(a+b)^3-3ab(a+b)$$
and
$$\cos^2{\frac{\theta}{2}=\frac{\cos\theta + 1}{2}}$$
Then I tried simplifying it:
$$\require{cancel} \begin{align} \sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}} & = (\sin^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}})^3 - 3\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}(\sin^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}})\\
& = 1 - 3\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}\\
& = 1 - 3(1 - \cos^2{\frac{\theta}{2}})\cos^2{\frac{\theta}{2}}\\
& = 1 - \frac{3}{2}(\cos\theta+1) + \frac{3}{4}(\cos\theta+1)^2\\
& = \frac{4\cancel{-6\cos\theta}-2+3\cos^2+3+\cancel{6\cos\theta}}{4}\\
& = \frac{1}{4}(5+3\cos^2\theta)\end{align} $$
I suspect I must have messed up with some sign somewhere. The trouble is, I can't seem to find where. Please help me in this regard.

Update: I am not accepting an answer because all the answers are equally good. It would be an injustice to the other answers.

Note to the editors: I also suspect that my post is a little short on appropriate tags. If required, please do the needful.

You may use
$$
2\sin x \cos x= \sin (2x) \tag1
$$ 
giving 
$$
\begin{align} \sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}} & = (\sin^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}})^3 - 3\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}(\sin^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}})\\
& = 1 - 3\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}} \\
& = 1 - \frac34\sin^2{\theta}\quad (\text{using} \,(1))\\
& = 1 - \frac34(1-\cos^2{\theta})\\
& = \frac14(1+3\cos^2{\theta})
\end{align}
$$ as desired.
