What is the remainder when m+n is divided by 1000 in a trigonometric challenge?

[anemone]

Let $x$ be a real number such that $\dfrac{\sin^4 x}{20}+\dfrac{\cos^4 x}{21}=\dfrac{1}{41}$. If the value of $\dfrac{\sin^6 x}{20^3}+\dfrac{\cos^6 x}{21^3}$ can be expressed as $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers, find the remainder when $m+n$ is divided by 1000.

 

[Opalg]

Spoiler: Easier than it looks!

If $\dfrac{\sin^4x}{20} + \dfrac{\cos^4x}{21} = \dfrac1{41}$ then $$21*41\sin^4x + 20*41(1 - \sin^2x)^2 = 20*21,$$ $$41^2\sin^4x - 2*20*41\sin^2x + 20^2 = 0,$$ $$(41\sin^2x - 20)^2 = 0.$$ Therefore $\sin^2x = \dfrac{20}{41}$, $\cos^2x = \dfrac{21}{41}$ and $$\dfrac{\sin^6x}{20^3} + \dfrac{\cos^6x}{21^3} = \dfrac1{41^3} + \dfrac1{41^3} = \dfrac2{68921}.$$ So $m = 2$, $n = 68921$, $m+n = 68923$ and the remainder when $m+n$ is divided by $1000$ is $923$.

 

[anemone]

Aww, very well done, Opalg! I have been trying to solve it for a number of times and for some reason, I didn't see the way to tackle it as you did! As always, thanks for your insightful solution!