$\frac{9a^4b^2-16a^2b^6}{3a^2b-4ab^2}$

$\int\left(\sin^3\left(3x\right)\cdot\cos\left(3x\right)\right)dx$

$k^2-24k-25=0$

$\frac{\tan\left(x\right)}{1+\cos\left(x\right)}+\frac{\tan\left(x\right)}{1-\cos\left(x\right)}=\frac{2}{\sin\left(x\right)\cos\left(x\right)}$

$20n^{11}+7n^7-6n^3$

$18c-\left(9a\right)$

$0.45c+c-0.89$