$\left(3y+4\right)^2\left(3y+4\right)^2$
$x^2+x-10=0$
$\left(10x^2+3y^6\right)^2$
$\left(x+2\right)\cdot\left(x-6\right)\cdot\left(x-1\right)$
$\left(27\sqrt[3]{z^2}\right)^2$
$5\left(x+6y\right)+3y-2x+1$
$2x^2-10x+8$
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