$\frac{d}{dx}x^3y^2-3x^4=8+y$
$\frac{x^2}{6}-5$
$10^5\cdot10$
$\left(x^{a+1}+x^{2a-1}\right)^3$
$\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)$
$3x^{2}+5x-2$
$6x^2+3a^2=12$
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