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.
 A: Let $\sin^2 \theta = k$
Then our equation is. 
$$\frac{k^2}{a}+\frac{(1-k^2)^2}{b}=\frac{1}{a+b}$$  
I won't solve the whole equation but i will put in some steps. 
We will be getting something messy like:
$$k^2ab+k^2b^2+a^2+a^2k^2-2a^2k^2+ab+abk^2-2abk=ab$$  
And the best part is this all simplifies to:  
$$(k\times (a+b)-a)^2=0$$ 
Hence $k=\frac{a}{a+b}=\sin^2 \theta$ 
$(1-k)=\frac{b}{a+b}=\cos^2 \theta$
Let us prove the general case. $$\frac{\sin^{4n}\theta}{a^{2n-1}}+\frac{\cos^{4n} \theta}{b^{2n-1}}=\frac{1}{(a+b)^{2n-1}}$$ 
Substituting our values on LHS.
$$\frac{1}{a^{2n-1}}\times \left(\frac{a}{a+b}\right)^{2n}+\frac{1}{b^{2n-1}}\times \left(\frac{b}{a+b}\right)^{2n}$$
$$\frac{a}{(a+b)^{2n}}+\frac{b}{(a+b)^{2n}}=\frac{1}{(a+b)^{2n-1}}=RHS$$ Hence result is true $\forall$ $n \in N$
A: 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.
