$$\int16\left(6x^2+8\right)dx$$

$$\left(4+\frac{3i}{5}\right)^3$$

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

$$\ln\left(x^4-4\right)=4\ln\left(x\right)-\ln\left(4\right)$$

$$5x^2-16x+3$$

$$2\sin x-\sqrt{3}=0$$

$$\left(-1-1\right)-\left(5-4+6+8\right)-\left(8-7\right)$$