$\left(6y^2-9y^3\right)^2$
$\left(-15\right)\cdot\left(-9\right)$
$tan\left(x\right)=\frac{1}{10}$
$2\cdot17-3$
$1+a=\frac{a^3+1}{a^2+1}$
$\left(150\right)+\left(-35\right)+\left(40\right)+\left(-15\right)$
$\left(b+c\right)^3$
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