$\left(x-3\left(x+3\right)\right)$
$3-1+5-9-3$
$y\:=\:\frac{sin^4x\cdot tan^6x}{\left(x^2+3\right)^2}$
$y^2=\:e^2x$
$\left(a^2+\:3\right)^3$
$3x-4y=2$
$3x.\left(-5x^4\right)$
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