$\left(x^3+2\right)\left(x^3-3\right)\left(x^3-1\right)$

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

$\frac{x^{-4}y^6y^3}{x^{-7}}$

$9a^2-6a^2x+3a^3x^2$

$3\sqrt{16a}-\sqrt[3]{54a}$

$x + 3 = \sqrt { 3 x - 1 }$

$y'=\frac{2-y}{x},\:y\left(3\right)=1$