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Explanation:

ab + ba = -1

⇒ a2 + b2 = -ab
⇒ a2 + b2 + ab = 0​​​​​​​   ...(1)

Now, we have 6(a3 - b3)

⇒ 6(a3 - b3) = (a - b)(a2 + b2 + ab)
⇒ 6(a3 - b3) = (a - b) × 0 = 0 [a2 + b2 + ab = 0 from (1)]

Hence, option (c).

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