I have the following polynomial square to solve
$(a+b(cos(2 \pi \nu )-sin(2 \pi \nu ))+c(cos(4 \pi \nu )-sin(4 \pi \nu )))^{2}$
For three factors I would use $(a+bx+cy)^{2} = (a^{2}+bx^2+cy^2+2bx+2cy)$ but in my case I have 5 factors and I don't know how to continue for a solution.
Thanks to all
Here, $x=2\pi \nu$. We have, $$\big( a+b(\cos x -\sin x) +c(\cos 2x -\sin 2x) \big)^2 \\ =a^2 +b^2(\cos x-\sin x)^2 +c^2(\cos 2x -\sin 2x)^2 +2ab(\cos x-\sin x) +2bc(\cos x-\sin x)(\cos 2x-\sin 2x) +2ac(\cos 2x-\sin 2x) \\ = a^2 +b^2(1-\sin2x) +c^2(1-\sin 4x)+2ab(\cos x-\sin x) +2bc\big( (\cos x \cos 2x +\sin x \sin 2x) -(\sin 2x \cos x +\sin x \cos 2x)\big)+2ac(\cos 2x-\sin 2x) \\ = a^2 +b^2(1-\sin 2x)+c^2(1-\sin 4x) +2ab(\cos x-\sin x)+2bc(\cos x-\sin 3x) +2ac(\cos 2x -\sin 2x) \\ \vdots $$ Hopefully you can do the simplification.