Discussion

Explanation:

Given, (r-s)3+(s-t)3+(t-r)36(r-s)(s-t)(t-r)

We know, if a + b + c = 0, then a3 + b+  c3 = 3abc

Here, (r - s) + (s - t) + (t - r) = 0

∴ (r - s)3 + (s - t)3 + (t - r)3 = 3(r - s)(s - t)(t - r)

∴ (r-s)3+(s-t)3+(t-r)36(r-s)(s-t)(t-r) = 3(r-s)(s-t)(t-r)6(r-s)(s-t)(t-r) = 12

Hence, option (b).

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