$\frac{3}{\frac{3}{3}}$

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

$\left(x-169\right)^2$

$\sqrt{\left(2^4\right)}\left(4^2\right)$

$3.\left(2-x\right)=3-2x$

$\frac{8x+6}{x+6}=\frac{5x-1}{x+2}$

$\left(x+1\right)^2\left(x-2\right)+\left(x+2\right)\left(x-1\right)\left(x+1\right)$