$\left(1+4ax\right)^3$
$4\left(-3\right)^3-3\left(-3\right)^2-36\left(-3\right)^2$
$-30\cdot\left(-2\right)\cdot0$
$f\left(x\right)=\left(2x-2\right)^2$
$\left(3\cdot10^{-1}\right)\left(0.6\right)^2$
$11.\left(3a+5b\right)\:\left(2b\right)$
$\left(2b^2+3c^4\right)^2$
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