$3a^3+15a^2$
$16x^2+4x+1;\:x=1$
$b^2-2b+3=0$
$f\left(x\right)=\left(x+4\right)^3\left(8x-9\right)^5$
$\int\frac{1}{x^2\left(x^2+3x+2\right)}dx$
$3ab,\:si\:a\:=\:1\:y\:b\:=\:2$
$\left(1+x\right)^{\frac{1}{3}}-1-\frac{x}{3}+\frac{x^2}{9}\le\:\frac{5}{8}x^3$
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