$x+8=3\left(x+4\right)$

$x^2+y^2=\left(x+yi\right)\left(x-yi\right)$

$\frac{\left(3a^3-5ab^2-6a^2b^3\right)}{2a}$

$36x^2-121$

$3\left(x^2-2x-5\right)^2+4\left(x^2-2x-5\right)-7$

$\left(8-11\right)-\left(3+1-4-6\right)$

$x=\frac{\sin\left(3x\right)}{\sin\left(x\right)}-\frac{\cos\left(3x\right)}{\cos\left(x\right)}$