$\left(x-\frac{1}{3}\right)^4$
$6x^2-8+5a-3x-\left(x\left(6x-12\right)+4x-8\right)$
$50\cdot12$
$\left(8x-10\right)\left(-4\right)$
$\frac{16x^2+4x}{16x^2+16x+3}$
$\:\frac{x^6+64}{x+2}$
$2x+2y+2y'-1=0$
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