$10x^5+7=40x+7$
$2\left(y+3\right)+6\left(y-4\right)+5$
$\left(1+\tan x\right)^2-\left(1-\tan x\right)^2$
$\left(2x+1\right)\left(3x+6\right)=0$
$\frac{\left(16x^2+16x+8\right)}{\left(3x+2\right)}$
$12w^5y^5-16w^4y^4+18w^3y^3$
$a^2-9a$
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