$\left(2x^3-3\right)\left(3x^3-3\right)$
$2x^2+4x-5=1$
$y'+10y\:=\:0\:y\left(0\right)\:=\:4\:$
$-0.5\left(-2\cdot5\right)$
$\left(\frac{2}{5}a^2+3b\right)\left(\frac{2}{5}a^2-3b\right)$
$x^3+3x^2+3x+1$
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