$3x^2-9x=-27$
$\left(6\:-\:17\right)\:.\:\left(-21\:+\:17\right)$
$\left(3x^3-6\right)\left(x^3-1\right)$
$x^2+9x+27$
$\left(\sqrt{x}+4\right)^2\:+\left(\sqrt{x}-4\right)^2$
$\frac{\sin^4\left(x\right)+\sin^2\left(x\right)\cdot\cos^2\left(x\right)}{1+\cos\left(x\right)}$
$100a^{2}b^{3}c-150ab^{2}c^{2}+50ab^{3}c^{3}$
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