$\left(-3y\right)-\left(-5y\right)+\left(+6y\right)$
$\left(1-\sin^2z\right)\cdot\left(1+\tan^2z\right)=1$
$\int\frac{\left(x+1\right)^2}{e^x}dx$
$y^2-6y+9=0$
$\arcsin\left(0.48\right)$
$\int\left(1-x^3\right)^2\left(x\right)dx$
$x^3y'+4x^2y=x$
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