$\frac{x^3+4x^2-3x-18}{x^2-3}$
$\left(3xy^3+9x-4x^2\right)+\left(8x^2+4xy^3-x\right)$
$\left(3a^4b^3\right)^6$
$8=u$
$4y^2-2y+1=0$
$2x^2+24xy+16y^2$
$x^2-4x+16=0$
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