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

Hint

Set $\sin^2x=s\iff\cos^2x=1-s$ in the given condition to form a quadratic equation in $s$

Solve to find $s=\sin^2x=\dfrac a{a+b}$

$\cos^2x=?$

We know, $~\sin^2x + \cos^2x = 1 ⇒\cos^2x = 1 - \sin^2x~$ and hence $~\cos^4x = (1 - \sin^2x)^2 = 1 + \sin^4x - 2\sin^2x~$.

Now

$~\dfrac{\sin^4x}{a}+\dfrac{\cos^4x}{b}=\dfrac{1}{a+b}~$
$~\implies\dfrac{\sin^4x}{a} + \dfrac 1b(1 + \sin^4x - 2\sin^2x) = \dfrac{1}{a + b}~$
$~\implies \dfrac{\sin^4x}{a} + \dfrac 1b + \dfrac{\sin^4x}{b} - \dfrac{2}{b}\sin^2x = \dfrac{1}{a + b}~$
$~\implies \sin^4x\left(\dfrac 1a + \dfrac 1b\right) - \dfrac{2}{b}\sin^2x = \dfrac{1}{a + b} - \dfrac 1b~$
$~\implies\dfrac{a+b}{ab}\sin^4x- 2a\dfrac{\sin^2x}{ab} = \dfrac{b - a - b}{(a + b)b}~$
$~\implies(a + b)^2(\sin^2x)^2 - 2a(a + b) \sin^2x = -a²~$
$~\implies \{(a + b)\sin^2x\}^2 -2.a.(a + b)\sin^2x + a^2 = 0~$
$~\implies \{(a + b)\sin^2x - a\}^2 = 0~$
$~\implies \sin^2x = \dfrac{a}{a + b}~$
$~⇒1 - \sin^2x = \cos^2x = 1 - \dfrac{a}{a + b} = \dfrac{b}{a + b}~$
Therefore, $~\sin^2x = \dfrac{a}{a + b}~$ and $~\cos^2x = \dfrac{b}{a + b}~$.

Now

$~\dfrac{\sin^6x}{a^2}+\dfrac{\cos^6x}{b^2}~$
$~=\dfrac 1{a^2}\left(\dfrac{a}{a + b}\right)^3+\dfrac 1{b^2}\left(\dfrac{b}{a + b}\right)^3~$
$~=\dfrac{a}{(a + b)^3}+\dfrac{b}{(a + b)^3}~$
$~=\dfrac{1}{(a + b)^2}~$