$\lim_{x\to0}\left(\frac{1-\cos\left(8x\right)}{x}\right)$

$2x^2-4xy-6y^2$

$\left(2b\:-\:4c\right)\left(2b\:+\:4c\right)\:$

$-5\cdot2\:4\cdot4$

$\frac{1-\left(\sin\left(x\right)\right)^4}{1+\:\left(\sin\left(x\right)\right)^2}=\left(\cos\left(x\right)\right)^2$

$2.\left(a^2\:+\:4\right)\left(a^2-4\right)$

$704b^3x^4y+256b^6+484x^8y^2$