$sin\left(3x\right)=sinx\left(4cos^2x-1\right)$
$\frac{\frac{2}{2x-1}-1}{4x-2+1}$
$-\frac{3x^{-2}}{y^3}$
$3r=\frac{dr}{dt}-6t^3$
$\left(\frac{1}{3}x+\frac{1}{3}y\right)\left(\frac{4}{5}x\right)$
$3x^3-13x^2+2x+8$
$\left(-200\right)\:-\:\left(-147\right)$
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