$-\frac{3}{x^2+8x+16}$

$\left(\frac{1}{2}y^2-6\right)^3$

$\lim_{x\to\infty}\left(\frac{1-x\left(x^3-6x^2\right)^{-\frac{1}{3}}}{\left(x^3-6x^2\right)^{\frac{-1}{3}}}\right)$

$\frac{x^3}{x^{-2}}^2$

$yx=3$

$\left(2x+9\right)\left(2x+7\right)$

$2\cos\left(x\right)-5=3\cos\left(x\right)-4$