$7x^2+4x^2+7x+2+6x^3+8x+3$
$\frac{x^4+5x^3+3x^2}{x}$
$\left(f^4-2\right)\left(f^4+5\right)$
$2\sqrt[3]{a^6}b$
$\left(8x+4\right)^4$
$7x-3^2$
$1+\frac{\tan^2\left(x\right)}{\sec\left(x\right)+1}=\sec\left(x\right)$
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