$\sqrt[3]{64^6}$
$\left(-\infty\right)^2+4$
$3.5x-3=0.5$
$s+\frac{1}{2}s^3+\frac{1}{5}s^7$
$\left(-19\right)\cdot\left(-10\right)$
$\frac{x^5+4x^3-7x^2+8}{\left(x-1\right)}$
$2+\sin\left(3x\right)=0$
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