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?

Recall that 
$$\sin^2\alpha = 1 - \cos^2\alpha$$
and express everything in terms of $\cos^2\alpha$:
$$\begin{align}
R &= \left(\;\sin^2\alpha\;\right)^3 + \left(\;\cos^2\alpha\;\right)^3+K\left(\;\left(\;\sin^2\alpha\;\right)^2+\left(\;\cos^2\alpha\;\right)^2\;\right) \\
&= \left(\;1-\cos^2\alpha\;\right)^3 + \left(\;\cos^2\alpha\;\right)^3+K\left(\;\left(\;1-\cos^2\alpha\;\right)^2+\left(\;\cos^2\alpha\;\right)^2\;\right) \\
&= \left(\;1-x\;\right)^3 + \left(\;x\;\right)^3+K\left(\;\left(\;1-x\;\right)^2+\left(\;x\;\right)^2\;\right) \qquad\text{(writing $x$ for $\cos^2\alpha$)}\\
&= 1 - 3 x + 3 x^2 + K \left(\; 1 - 2 x + 2 x^2 \;\right) \\
&= 1 +K -(3+2K) x + (3+2K) x^2
\end{align}$$
Independence from $\alpha$ translates to independence from $x$. We need a value of $K$ that causes the non-constant terms of the polynomial to vanish. Clearly, $K = -3/2$. $\square$
