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The term independent of $x$ in the expansion of $\left(\frac{1}{60} - \frac{x^8}{81}\right) \cdot \left(2x^2 - \frac{3}{x^2}\right)^6$ is equal to:
(1) $-72$ (2) 36 (3) $-36$ (4) $-108$
$= \frac{-4320}{60} + 36 = -72 + 36 = -36$.
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