$3x^4-48$
$\:\:+1-\left(-8-10\right)+\left(+7-5\right)$
$3y^2+5y=12$
$\frac{3}{4}\:m^3\:y,\:m=4,\:y=11$
$x^4+8x^3+2x^2-80x-75$
$y=4x^2-3x+5$
$( q ^ { s } ) ^ { 3 }$
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