If $\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}$, then show that $\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}$ 
If $\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}$, then show that $\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}$  

My work:
$(\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b})=\frac{1}{a+b}$
By squaring both sides, we get,
$\frac{\sin^8 x}{a^2}+\frac{\cos^8 x}{b^2}+2\frac{\sin^4 x \cos^4 x}{ab}=\frac{1}{(a+b)^2}$
$\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}-2\frac{\sin^4 x \cos^4 x}{ab}-\frac{\sin^6 x \cos^2 x}{a^2}-\frac{\sin^2 x \cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
So, now, we have to prove that,
$-2\frac{\sin^4 x \cos^4 x}{ab}-\frac{\sin^6 x \cos^2 x}{a^2}-\frac{\sin^2 x \cos^6 x}{b^2}=0$
I cannot do this. Please help!

One's first inclination might be to reduce powers by invoking the Half-Angle Identities, which can be written as 
$$\sin^2 x = \frac{1}{2}( 1 - k ) \qquad \cos^2 x = \frac{1}{2}(1 + k) \qquad\text{, where}\qquad k := \cos 2x$$
Then,
$$\begin{align}
\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b} = \frac{1}{a+b} \quad
&\implies\quad \frac{(1-k)^2}{4a}+\frac{(1+k)^2}{4b}=\frac{1}{a+b} \\[4pt]
&\implies\quad \big(\;k\;(a+b)+a-b\;\big)^2 = 0 \\[4pt]
&\implies\quad k = -\frac{a-b}{a+b} \\[4pt]
&\implies\quad \sin^2 x = \frac{a}{a+b} \quad\text{and}\quad \cos^2 x = \frac{b}{a+b}
\end{align}$$
Therefore,
$$\frac{\sin^6 x}{a^2} + \frac{\cos^6 x}{b^2} = \frac{1}{a^2}\left(\frac{a}{a+b}\right)^3 + \frac{1}{b^2}\left(\frac{b}{a+b}\right)^3 = \frac{a+b}{(a+b)^3} = \frac{1}{(a+b)^2}$$
