$\left(2x^2-5x+6\right)\cdot\left(3x^2-4x+5\right)$
$-18\cdot15$
$\left(p+4\right)^4$
$4-\left(10-6+\left(-3\right)-6\right)+\left(3+4+\left(-2\right)-\left(-4\right)\right)$
$\left(4x+6y\right)\left(4x-14y\right)$
$x^4+4y^4-3=0$
$\left(4p-3q\right)^3$
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