$5\left(3x-4\right)=5\left(3x+4\right)$
$\frac{1}{3}x^3+4x^2+x$
$\left(x^2+2\right)\left(x-1\right)^3$
$\frac{2x^2}{\left(x+y\right)^2}$
$ab^2+ba^2$
$\left(y^{-1}+1\right)dx-xy^{-2}dy=0$
$-12,43+7,41$
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