$\left(3xy^{-4}\right)^{-2}$
$x^2+16x+40$
$\frac{\left(6x\left(2\right)+16x+8\right)}{\left(3x+2\right)}$
$\left(8x^2\right)\left(x+3\right)$
$5x+64=14$
$8+4\left(x+8\right)<10+x$
$43+38c+35c-21c-32c$
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