Prove that: $$\dfrac{1}{2^{180}a^{360}}\dfrac{(a^{720}-1)(a^2-1)}{a^{2}+1} = \dfrac{\left(1+\dfrac{\sqrt{3}}{2}\right)^{180} - \left(1-\dfrac{\sqrt{3}}{2}\right)^{180}}{\sqrt{3}}$$ where: $$a = \dfrac{1+\sqrt{3}}{\sqrt{2}}.$$
This seemed like a very challenging question but the fact that we have a telescoping binomial sum in the numerator of the LHS helps. I think if we can simplify the LHS sufficiently we might be able to prove it by just equating both sides of the equation.
HINT:
Observe that $a^2=2+\sqrt3$
$\implies1+\dfrac{\sqrt3}2=\dfrac{a^2}2$ and $1-\dfrac{\sqrt3}2=\dfrac1{2a^2}$
Use Componendo and Dividendo to find $\dfrac{a^2-1}{a^2+1}=?$
Now
$$\dfrac{1}{2^{180}a^{360}}\dfrac{(a^{720}-1)(a^2-1)}{a^{2}+1}=\left[\left(\dfrac{a^2}2\right)^{180}-\left(\dfrac1{2a^2}\right)^{180}\right]\cdot\dfrac{a^2-1}{a^2+1}$$
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