$x+\left(x+3\right)\left(x-3\right)=3+x\left(x+1\right)$
$\frac{x^2-\frac{1}{4}x-3}{x+2}$
$5a^{-1-2}$
$\left(4ba^4\right)^2$
$-a+b+2b+2c+3a+2c+3b$
$\frac{-1^3-1}{-1^2}$
$z^2+z+0$
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