$$\left(x-1\right)^2\le\left(x+1\right)^2+4$$

$$\left(+14\right)-\left(-12\right)$$

$$\left(y-8\right)\left(y+8\right)$$

$$\frac{\left(\sec\left(x^3+x^2\right)\right)^4}{\cos\left(x\right)}$$

$$x^5\left(8-x\right)$$

$$\frac{x^3}{16x^4}$$