$\int\left(\frac{1}{\sqrt{x}\left(1+\sqrt[3]{x}\right)^2}\right)dx$

$_{x\to\:-3}\frac{x^3+x^2-6x}{x^2+3x}$

$\left(\frac{x}{2}+\frac{5y}{4}\right)\cdot\left(\frac{x}{2}-\frac{5y}{4}\right)$

$\left(9\right)\left(-2\right)\left(4\right)$

$7x\:-\:3\:-\:\frac{1}{2}\left(6x\:+\:4\right)$

$15x^2-14x=8$

$\left(-\frac{2}{3}x^2+\frac{1}{5}x-3\right)\left(\frac{3}{4}x^2-x+\frac{3}{2}\right)$