$\left(3-3\right)^2+5\left(3\right)$
$\frac{5x-6}{2}>x+2$
$f\left(x\right)=\left(5x^4+3x^2-4\right)^3$
$y^4-3x^2-4=0$
$\frac{6x^3+24x^2+18x}{3x}$
$9a^3b^2+15a^2b^2c+21ab^2$
$f\left(x\right)=\frac{x^2\left(x-4\right)^3}{x+1}$
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