$\left(\frac{1}{3}x^3\right)^3$
$-\frac{x^3}{3}+x^2+8x;\:x=-2;\:y=0$
$\frac{\frac{1}{x}}{x}$
$\cos\left(x\right)=\cos\left(2x\right)+\cos\left(4x\right)$
$x^2+10x-28$
$\frac{45x^2-17x-48}{14x^3-5x^2-24x}$
$\frac{2}{7}x+8=\frac{2}{3}x$
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