$3\sin\left(3x\right)+3$
$\frac{4x^2-8x+3}{36x^3+66x^2+40}$
$\lim_{x\to\infty}\left(\frac{11}{3x+9x^5}\right)$
$+32b+60b-20b+96c-10c-9c$
$\left(-10x+4\right)^2$
$\frac{160x\:-36}{5x+3}$
$\sqrt{\frac{6ab^3c}{2a^3}}$
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