Simplifying
$3x^3-xy^3=8$
$5x-2y=1$
$4x-2>2x+2$
$\frac{1}{2}a-\frac{3}{4}a+\frac{3}{8}a$
$\lim_{t\to\infty}\left(\frac{\left(t+2\right)^{\frac{3}{8}}}{t^{\frac{3}{4}}}\right)$
$\lim_{x\to16}\left(\frac{4-\sqrt[2]{x}}{16-x^2}\right)$
$-yx^2z^3\left(4y^0z\right)^2$
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