If si$n^4$θ/a + co$s^4$θ/$b=1/(a+b)$, where a and $b$ are positive, then si$n^8$θ/$a^3$ + co$s^8$θ/$b^3$ equals:

Put $x$ = si$n^2$θ. Clearing denominators turns the given equation into $((a+b)x-a{)}^2=0$, so the only possibility is si$n^2$θ = $a/(a+b)$ and co$s^2$θ = $b/(a+b)$. Then si$n^8$θ/$a^3=[a^4/(a+b{)}^4]/a^3=a/(a+b{)}^4$ and co$s^8$θ/$b^3=b/(a+b{)}^4$, and their sum is $(a+b)/(a+b{)}^4=1/(a+b{)}^3$. Check with $a=2,b=3$: si$n^2$θ = 0.4 makes the given side 0.2 = 1/5 and the target side 0.008 = 1/125. Each power of the ratio raises the exponent by one, so $1/(a+b{)}^2$ is one step short and $1/(a^3+b^3)$ cubes the wrong thing.