$\lim_{x,y\to0,0}\left(\frac{x^3-y^3}{x^2+y^2}\right)$
$3x^2+xy^3=0$
$\left(-8b-2\right)\left(b+9\right)\left(2b+3\right)$
$\left(6a^2+5\right)\left(6a^2+7\right)$
$\left(2-3x\right)\left(4-x\right)$
$-8+-+-4+-5$
$2^6\cdot2^3$
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