Problem: If $\frac{\sin^4\theta}{x}+\frac{\cos^4\theta}{y}=\frac{1}{x+y}$, show that, $\frac{\sin^{2m+2}\theta}{x^m}+\frac{\cos^{2m+2}\theta}{y^m}=\frac{1}{(x+y)^m}$.
My attempt: I observed that if both the sides of the given equation are raised to the power of m, the R.H.S of the resulting equation matches with the R.H.S of the result that has to be proved. After that we need to simplify the L.H.S of the resulting expression to match it with the L.H.S of the result that has to be proved.
$\frac{\sin^4\theta}{x}+\frac{\cos^4\theta}{y}=\frac{1}{x+y}$
$\implies(\frac{\sin^4\theta}{x}+\frac{\cos^4\theta}{y})^m=(\frac{1}{x+y})^m$
$\implies(\frac{y\sin^4\theta+x\cos^4\theta}{xy})^m=\frac{1}{(x+y)^m}$
$\implies\frac{(y\sin^4\theta+x\cos^4\theta)^m}{x^my^m}=\frac{1}{(x+y)^m}$
My problem: I am stuck at this step and cannot understand how to proceed, that is, how to simplify $(y\sin^4\theta+x\cos^4\theta)^m$. Please help. A continuation of my method as well as alternate methods (except mathematical induction) are welcome.
The condition gives $$\frac{\sin^4\theta}{x}+\frac{\cos^4\theta}{y}=\frac{(\sin^2\theta+\cos^2\theta)^2}{x+y}$$ or $$\left(\frac{1}{x}-\frac{1}{x+y}\right)\sin^4\theta-\frac{2\sin^2\theta\cos^2\theta}{x+y}+\left(\frac{1}{y}-\frac{1}{x+y}\right)\cos^4\theta=0$$ or $$\left(\frac{\sin^2\theta}{x}-\frac{\cos^2\theta}{y}\right)^2=0,$$ which gives $$\frac{x}{y}=\tan^2\theta.$$ Thus, we need to prove that $$\sin^{2m+2}\theta+\left(\frac{x}{y}\right)^m\cos^{2m+2}\theta=\left(\frac{\frac{x}{y}}{\frac{x}{y}+1}\right)^m$$ or $$\sin^{2m+2}\theta+\tan^{2m}\theta\cos^{2m+2}\theta=\left(\frac{\tan^2\theta}{\tan^2\theta+1}\right)^m,$$ which is $$\sin^{2m}\theta(\sin^2\theta+\cos^2\theta)=\sin^{2m}\theta.$$ Done!