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?

Let $x = \sin^2 \alpha,y = \cos^2 \alpha$. Note that $x+y = 1$ so $x^3 + y^3 = (x+y)(x^2+y^2-xy) = (x+y)((x+y)^2-3xy) = 1-3xy$
Also $x^2+y^2 = (x+y)^2-2xy = 1-2xy$
$R = (1-3xy) + K(1-2xy) = 1+K - (3+2K)xy$. So if $K = -3/2$ then $R$ will be independent of $\alpha$. If $K \neq -3/2$ then it will depend on $xy$ which then depends on $\alpha$.
Generally when you see an expression in the form of $\sin^{2n}x + \cos^{2n}x$ it helps to factor out $\sin^2 + \cos^2$ like what I did above
