$3\left(\frac{9}{16}a^2\left(\frac{4}{3}b\right)\right)$
$\frac{1}{\sec\left(b\right)-\tan\left(b\right)}=\frac{1+\sin\left(b\right)}{\cos\left(b\right)}$
$x^5y^5=x^5y^5$
$\frac{32x^2y^3}{2x}$
$3-6x>0$
$\int e^{\sqrt{3u+9}}du$
$2x^3\cdot5x^2$
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