Discussion

Explanation:

Given, $\frac{{(r-s)}^3+{(s-t)}^3+{(t-r)}^3}{6(r-s)(s-t)(t-r)}$

We know, if a + b + c = 0, then a³ + b³ +  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)

∴ $\frac{{(r-s)}^3+{(s-t)}^3+{(t-r)}^3}{6(r-s)(s-t)(t-r)}$ = $\frac{3(r-s)(s-t)(t-r)}{6(r-s)(s-t)(t-r)}$ = $\frac{1}{2}$

Hence, option (b).

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