Trigonometry Proof: Proving Sin^8 + Cos^8 = (a+b)^3

[The legend]

Homework Statement

Given the equality
$$\frac{sin^4(x)}{a} + \frac{cos^4}{b} = \frac{1}{a+b}$$

Prove that :
$$\frac{sin^8(x)}{a^3} + \frac{cos^8}{b^3} = \frac{1}{(a+b)^3}$$

The Attempt at a Solution

I cubed on both the sides of the 1st equation and solved a bit, reaching no where. Then I tried by cross multiplying a+b, getting
$$sin^4(x)+ cos^4(x) + \frac{b}{a}sin^4x + \frac{a}{b}cos^4x = 1$$

$$\frac{b}{a}sin^4x + \frac{a}{b}cos^4x = 2sin\frac{x}{2}cos\frac{x}{2}$$

Cubing this one didnt seem appropriate either. Maybe this is the wrong way I'm going in
Please help..

 

[iRaid]

I think you can cube the 2nd equation :p