$3x^3y'+4x^2=x^3$
$a^3-7a-6$
$-2x^4+4x^3+6x^2-8x+8$
$r\left(x-4\right)$
$\frac{1}{\cos\left(x\right)}-\frac{\cos\cot\left(x\right)\left(x\right)}{1+\cos\left(x\right)}$
$\frac{d}{dx}y=\sqrt[3]{x}+\frac{1}{\sqrt[3]{x}}$
$4\cos^2a=0$
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