$6\left(\frac{1}{3}\right)^{\infty-1}$

$b.-9x+8\:=\:5x\:-\:6$

$\left(\sec\left(a\right)\right)^3\left(\cos\left(a\right)\right)^2$

$\int\left(x+1\right)\:\left(x-2\right)^3dx$

$x^2-2x+10=0$

$\left(4xb^2-5xy^3\right)^2$

$\int_1^{e^2}\left(\frac{1}{x^{\sqrt{2}}}\right)dx$