$2+3-5+4-6+3-8-2+4-2$

$\sec\left(x\right)-\left(\tan\left(x\right)\sin\left(x\right)\right)=\frac{1}{\sec\left(x\right)}$

$\left(4x^6+6y^5-2z\right)^2$

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

$3^2-5^2$

$x^2+\frac{3}{2}x+\frac{9}{6}$

$\left(3b^{2}+2c^{3}\right)\left(3b^{2}-2c^{3}\right)$