$\left(10x^2+32x-45\right)-\left(x^3-6x^2+32x\right)$

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

$\left(4x^2+10\right)\left(4x^2+2\right)$

$5a^2+55a$

$\left(-2x^2y^3\cdot\frac{4xy^2}{3y^4}\right)^4$

$-50\cdot16$

$25a^2-15a-9$