$\left(\frac{0.05562}{-0.156}\right)-1\cdot100$
$\left(16mn\right)\left(8m^2n^2\right)$
$\left(x-2y^2\right)\left(x+4y^{-2}\right)$
$60\cdot16$
$\left(y+8\right)^3$
$\lim_{x\to0}\frac{\left(sen\left(2x\right)\:sen\left(3x\right)\right)}{\left(x-x^3\right)^2}$
$\left(4m+12n\right)\left(4m-12n\right)$
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