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.
 A: $$\sin^6\frac{\theta}{2}+\cos^6\frac{\theta}{2}=\cdots=1-3\left(1-\cos^2\frac{\theta}{2}\right)\cos^2\frac{\theta}{2}$$
is correct. From here, note that 
$$\begin{align}\sin^6\frac{\theta}{2}+\cos^6\frac{\theta}{2}&=1-3\left(1-\color{red}{\cos^2\frac{\theta}{2}}\right)\color{red}{\cos^2\frac{\theta}{2}}\\&=1-3\left(1-\color{red}{\frac{\cos\theta+1}{2}}\right)\cdot\color{red}{\frac{\cos\theta+1}{2}}\\&=1-3\cdot\frac{1-\cos\theta}{2}\cdot\frac{1+\cos\theta}{2}\\&=\frac{4-3(1-\cos^2\theta)}{4}\\&=\frac{1+3\cos^2\theta}{4}\end{align}$$
A: You may begin factoring:
\begin{align*}
\sin^6\frac\theta2+\cos^6\frac\theta2 &=\Bigl(\sin^2\frac\theta2+\cos^2\frac\theta2\Bigr)\Bigl(\sin^4\frac\theta2-\sin^2\frac\theta2\cos^2\frac\theta2+\cos^4\frac\theta2\Bigr)\\
&=\Bigl(\cos^2\frac\theta2-\sin^2\frac\theta2\Bigr)^2+\sin^2\frac\theta2\cos^2\frac\theta2=\cos^2\theta+\frac14\sin^2\theta\\
&=\frac14(3\cos^2\theta+1).
\end{align*}
You also may linearise:
$$\cos^2\theta+\frac14\sin^2\theta=\frac12(1+\cos2\theta)+\frac18(1-\cos2\theta)=\frac18(3\cos2\theta+5).$$
A: 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.
A: Hint: $$\sin^2\theta/2\cos^2\theta/2\\=\frac{1}{4}(1-\cos\theta)(1+\cos\theta)=\frac{1}{4}(1-\cos^2\theta)$$
A: A little life-saving trick:
$$
\require{cancel} \begin{align} \sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}} & = (\sin^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}})^3 - 3\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}(\sin^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}})\\
& = 1 - 3\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}\\
& = 1 - 3(1 - \cos^2{\frac{\theta}{2}})\cos^2{\frac{\theta}{2}}\\
& = 1 - \frac{3}{2}(\cos\theta+1) + \frac{3}{4}(\cos\theta+1)^2\\
& = \frac{4\cancel{-6\cos\theta}-2+3\cos^2+3+\cancel{6\cos\theta}}{4}\\
& = \frac{1}{4}(5+3\cos^2\theta)\end{align}
$$
is wrong because, if you evaluate at $\theta=0$ you get
$$
\require{cancel} \begin{align} 1 & = 1\\
& = 1 \\
& = 1 \\
& = 1 \\
& = 2\\
& = 2\end{align}
$$
and this also shows the passage that is wrong.
A: Let $\theta/2=x$. Then the LHS is:
$$\sin^6 x+\cos^6 x=\underbrace{\left(\sin^2 x+\cos^2 x\right)}_{=1}\left(\sin^4 x-\sin^2 x\cos^2 x+\cos^4 x\right)$$
$$=\left(1-\cos^2 x\right)^2-\left(1-\cos^2 x\right)\cos^2 x+\cos^4 x\tag{1}$$
The RHS is:
$$\frac{1}{4}\left(1+3\cos^2 2x\right)=\frac{1}{4}\left(1+3\left(2\cos^2 x-1\right)^2\right)\tag{2}$$
Now prove $(1)=(2)$.
