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!
 A: HINT:
Use $$\cos^2x=1-\sin^2x$$ to form a Quadratic Equation in $\displaystyle\sin^2x$ 
writing $\displaystyle\sin^2x=p$  we get $$\frac{p^2}a+\frac{(1-p)^2}b=\frac1{a+b}$$
$$\implies b p^2+ a(1+p^2-2p)=\frac{ab}{a+b}$$
$$\implies (a+b)\{(a+b)p^2-2ap+a\}=ab$$
$$\implies (a+b)^2p^2-2a\cdot (a+b)p+a^2=0\implies \left[p(a+b)-a\right]^2\implies p=\frac a{a+b}$$
Can you take it from here?
A: 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,
$$\frac{\sin^6 x}{a^2} + \frac{\cos^6 x}{b^2} = \frac{1}{a^2}\left(\frac{a}{a+b}\right)^3 + \frac{1}{b^2}\left(\frac{b}{a+b}\right)^3 = \frac{a+b}{(a+b)^3} = \frac{1}{(a+b)^2}$$
A: first we square both sides
$\dfrac{1}{(a+b)^2}=\dfrac{\sin^8x}{a^2}+2\dfrac{\sin^4x\cos^4x}{ab}+\dfrac{\cos^8x}{b^2}$
$=\dfrac{\sin^6x(1-\cos^2x)}{a^2}+2\dfrac{\sin^4x\cos^4x}{ab}+\dfrac{\cos^6x(1-\sin^2x)}{b^2}$
$=\dfrac{\sin^6x}{a^2}+\dfrac{\cos^6x}{b^2}-\dfrac{\sin^6x\cos ^2x}{a^2}-\dfrac{\cos^6x\sin^2x}{b^2}+2\dfrac{\sin^4x\cos^4x}{ab}$
it is sufficient to show the last three terms are zero
$=-\dfrac{\sin^6x\cos ^2x}{a^2}-\dfrac{\cos^6x\sin^2x}{b^2}+2\dfrac{\sin^4x\cos^4x}{ab}\\=-\sin ^2x\cos^2x(\dfrac{\sin^4x}{a^2}-\dfrac{2\sin^2 x\cos^2x}{ab}+\dfrac{\cos^4x}{b^2})$
$=-\sin ^2x\cos^2x(\dfrac{\sin^2x}{a}-\dfrac{\cos^2x}{b})^2$
Now consider the initial problem substituting $1=\cos^2x+\sin^2x$
we arrive to 
$\dfrac{\sin^4 x}{a}+\dfrac{\cos^4 x}{b}=\dfrac{\sin^2x+\cos^2x}{a+b}$
$\dfrac{\sin^4x}{a}-\dfrac{\sin^2x}{a+b}+\dfrac{\cos^4x}{b}-\dfrac{\cos^2x}
{a+b}=0$
$\dfrac{a\sin^4x+b\sin^4x-a\sin^2x}{a(a+b)}+\dfrac{b\cos^4x+a\cos^4x-b\cos^2x}{b(a+b)}=0$
factorise $a\sin ^2x $ and $b\cos^2x$ to get to
$\dfrac{a\sin^2x(\sin^2x-1)+b\sin^4x}{a}+\dfrac{b\cos^2x(\cos^2x-1)+a\cos^4x}{b}=0$
$\sin^2x(-\cos^2x)+\dfrac{b}{a}\sin^4x+\cos^2x(-\sin^2x)+\dfrac{a}{b}\cos^4x=0\\ \dfrac{b}{a}\sin^4x-2\sin^2x(\cos^2x)+\dfrac{a}{b}\cos^4x=0$
divide by $ab$ here we suppose $a,b\ne 0$
$\dfrac{1}{a^2}\sin^4x-\frac{2}{ab}\sin^2x(\cos^2x)+\dfrac{1}{b^2}\cos^4x=0$
$\dfrac{\sin^4x}{a^2}-\dfrac{2\sin^2 x\cos^2x}{ab}+\dfrac{\cos^4x}{b^2}=(\dfrac{\sin^2x}{a}-\dfrac{\cos^2x}{b})^2=0$
$a\ne -b$
$\fbox{}$
A: $\cos^4x=(\cos^2x)^2=(1-\sin^2x)^2=1+ \sin^4x-2\sin^2x$
Now,
$\dfrac{\sin^4x}{a} +\dfrac{cos^4x}{b}=\dfrac{1}{a+b}$
$\Rightarrow \dfrac{\sin^4x}{a} + \dfrac{(1+\sin^4x-2\sin^2x)}{b} = \dfrac{1}{a+b}$
$\dfrac{[b\cdot\sin^4x + a(\sin^4x-2\sin^2x+1)]}{ab} = \dfrac{1}{a+b}$
$\Rightarrow \dfrac{[(a+b)\sin^4x-2a \sin^2x+a]}{ab} = \dfrac{1}{a+b}$
$\Rightarrow (a+b)^2 \sin^4x - 2a(a+b)\sin^2x + a(a+b) =ab$
$\Rightarrow (a+b)^2 \sin^4x - 2a(a+b)\sin^2x + a^2=0$
$\Rightarrow [(a+b)\sin^2x-a]^2 = 0$
$\Rightarrow (a+b)\sin^2x - a = 0$
$\Rightarrow \sin^2x=\dfrac{a}{(a+b)}$
Taking third power on both side
$\sin^6x=\dfrac{a^3}{(a+b)^3}$
Dividing by both sides by $a^2$
$\dfrac{\sin^6x}{a^2}=\dfrac{a}{(a+b)^3}$
Now,
$\cos^2x=1 - \sin^2x=1-\dfrac{a}{(a+b)}=\dfrac{b}{(a+b)}$
$\Rightarrow \cos^2x=\dfrac{b}{(a+b)}$
Taking third power of both side
$\cos^6x=\dfrac{b^3}{(a+b)^3}$
Dividing both sides by $b^2$,
$\dfrac{\cos^6x}{b^2} =\dfrac{b}{(a+b)^3}$
$\Rightarrow \dfrac{\sin^6x}{a^2} + \dfrac{\cos^6x}{b^2}=\dfrac{a}{(a+b)^3} + \dfrac{b}{(a+b)^3} =\dfrac{(a+b)}{(a+b)^3}=\dfrac{1}{(a+b)^2}$ 
Hence proved.
A: My Solution:: Given $$\displaystyle \frac{\sin^4 x}{a}+\frac{\cos^4 x}{b} = \frac{1}{a+b}.$$ 
Now using the Cauchy-Schwarz inequality we get
$$\displaystyle \frac{(\sin^2 x)^2}{a}+\frac{(\cos^2 x)}{b}\geq \frac{\left(\sin^2 x+\cos^2 x\right)^2}{a+b} = \frac{1}{a+b}$$
and equality holds when $$\displaystyle \frac{\sin^2 x}{a} = \frac{\cos^2 x}{b}.$$
Now using ratio and proportion, we get $$\displaystyle \frac{\sin^2 x}{a} = \frac{\cos^2 x}{b}=\frac{\sin^2 x+\cos^2 x}{a+b}=\frac{1}{a+b}.$$
So $$\displaystyle \sin^2 x = \frac{a}{a+b}$$ and $$\displaystyle \cos^2 x = \frac{b}{a+b}.$$ So we get $$\displaystyle \frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{a^3}{a^2\cdot (a+b)^3}+\frac{b^3}{b^2\cdot (a+b)^3} = \frac{1}{(a+b)^2}$$
