If $\frac {\sin^4\theta}a +\frac {\cos^4\theta}b=\frac 1 {a+b}$ P.T $\frac {\sin^8\theta}{a^3} + \frac {\cos^8\theta} {b^3} = \frac1{(a+b)^3}$ The question reads if
$$\frac {\sin^4\theta}  {a} + \frac {\cos^4\theta}  {b} = \frac 1 {a+b}$$ 
 Then prove that 
$$\frac {\sin^8\theta}  {a^3} + \frac {\cos^8\theta}  {b^3} = \frac 1 {(a+b)^3}$$ 
I know simply cubing the first equation would not work. I have also tried arriving at the solution using various identities like $ \sin^2\theta + \cos^2\theta = 1$ but to no avail. I would also like to know the approach one should use while solving such trigonometry questions. 
I am an eleventh grader so sorry if this question is too silly.

Let $\frac{\sin^4\theta}{a}=x$ and $\frac{\cos^4\theta}{b}=y$ 
Cubing $(x+y)$
$(x+y)^3=x^3+y^3+3xy(x+y)$ 
Substituting.. 
$\frac1{(a+b)^3}=\frac{\sin^{12}\theta}{a^3}+\frac{\cos^{12}\theta}{b^3}+3\frac{\sin^4\theta\cos^4\theta}{ab}(\frac{\sin^4\theta}{a}+\frac{\cos^4\theta}{b})$ 

Now equate this expression to $\frac {\sin^8\theta}{a^3} + \frac {\cos^8\theta} {b^3}$ And LHS=RHS. Hence proved.
