$\left(4m^5\sqrt[3]{2}+3m\sqrt[3]{4}\right)^2$
$\frac{x^2}{64}-\frac{y^2}{36}=1$
$4v^4\cdot4v^4$
$\left(m^2+\frac{2}{7}\right)^3$
$f\left(x\right)=4x^2-4x-3$
$98^2$
$\left(\sqrt[3]{x+h}\right)^2$
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