$\frac{\sin2x+\sin6x}{\cos2x+\cos6x}=\tan\left(4x\right)$

$\int u\sqrt{\left(1-u^2\right)}du$

$6\cdot6\cdot6\cdot n^3\cdot n^3\cdot n^3\cdot b^4\cdot b^4\cdot b^4$

$2-\left(10\:\cdot3-6\right)$

$a^4\:+\:b^4\:=\:\left(a+b\right)\left(a^3-a^2b-ab^2+b^3\right)$

$\left(x-y\right)xy$

$-\left(12\right)-\left(3\right)-\left\{-\left(+7\right)-\left(-6\right)-\left[\left(-12\right)-\left(+9\right)\right]-\left(23\right)\right\}$