How do you simplify an expression involving fourth and higher order trigonometric functions? The problem is as follows:
Which value of $K$ has to be in order that $R$ becomes independent from $\alpha$?.
$$R=\sin^6\alpha +\cos^6\alpha +K(\sin^4\alpha +\cos^4\alpha )$$
So far I've only come up with the idea that the solution may involve $R=0$, therefore
$$\sin^6\alpha +\cos^6\alpha +K(\sin^4\alpha +\cos^4\alpha)=0$$ 
as a result the expression becomes $0$ thus independent from $\alpha$, however the result is like this
$$-K=\frac{\sin^6\alpha +\cos^6\alpha}{\sin^4\alpha +\cos^4\alpha}$$
I am not sure if this is the right way.
Moreover, how can I simplify this expression, as it has order four and six?

Using $ \sin^2 \alpha + \cos^2 \alpha =1$ 
\begin{eqnarray*}
\sin^4 \alpha + \cos^4 \alpha =(\sin^2 \alpha + \cos^2 \alpha)^2 -2\sin^2 \alpha  \cos^2 \alpha = 1-2\sin^2 \alpha  \cos^2 \alpha \\
\sin^6 \alpha + \cos^6 \alpha =(\sin^2 \alpha + \cos^2 \alpha)(\sin^4 \alpha  -\sin^2 \alpha \cos^2 \alpha+ \cos^4 \alpha)  = 1-3\sin^2 \alpha  \cos^2 \alpha. \\
\end{eqnarray*}
So your equation can be simplified to
\begin{eqnarray*}
R=\sin^6\alpha +\cos^6\alpha +K(\sin^4\alpha +\cos^4\alpha ) \\
=1-3\sin^2 \alpha  \cos^2 \alpha +K(1-2\sin^2 \alpha  \cos^2 \alpha) \\
=1+K-(2K+3)\sin^2 \alpha  \cos^2 \alpha \\
\end{eqnarray*}
So it is independent of $\alpha$ when $\color{blue}{2K+3=0}$. (Giving the value $K=\color{red}{-\frac{3}{2}}$)
