Proving trigonometrical identity from given identity. 
Given $$\dfrac{\sin^4 A}{a}+\dfrac{\cos^4 A}{b}=\dfrac{1}{a+b}$$ prove that:$$\dfrac{\sin^8 A}{a^3}+\dfrac{\cos^8 A}{b^3}=\dfrac{1}{(a+b)^3}$$

Using given, I proved: $b\sin^2 A=a\cos^2 A$ Help me proceeding after this.

You are almost finished. From where you left off, let us say $$b \sin^2 A =a \cos^2 A \iff \frac{\sin^2 A}{a}=\frac{\cos^2 A}{b}=c$$
For some $c$. Thus $$\sin^2 A+\cos^2 A=1 \iff ac+bc=1 \iff c=\frac{1}{a+b} \tag{1}$$
Now, note $$\dfrac{\sin^8 A}{a^3}+\dfrac{\cos^8 A}{b^3}=\frac{a^4c^4}{a^3}+\frac{b^4c^4}{b^3}=ac^4+bc^4=c^4(a+b)=\frac{1}{(a+b)^3}$$
From $(1)$. We are done. 
