$-1+6\left(0-6\right)$
$x^6+2x^4+x^2$
$\int\frac{6x+2}{\left(2x+1\right)\left(x^2+3\right)}dx$
$12x^2-9x=0$
$-\sqrt{9}-\left(-4\right).\left(\sqrt{64}\right)-5.\left(-2\right)$
$x\left(x^2+x+1\right)^4$
$\:-\left(12\:-\:3\:-1\right)\:$
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