$sin^4x=\frac{3}{8}+\frac{1}{8}\left(cos\left(4x\right)\right)-\frac{1}{2}\left(cos\left(2x\right)\right)$
$cosx-1=-cosx$
$-2\cdot15$
$8m^6+n^3$
$8y-5\left(3y+2\right)$
$\left(y^{-12}\right)^{-9}$
$16m^4n^3-20m^3n^2-36m^2nz$
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