2002 AMC 10B Problem 11

Let the integers be $n - 1,$ $n,$ $n + 1.$ Their product is $n(n^2 - 1)$ and their sum is $3n,$ so $$n(n^2 - 1) = 8(3n) = 24n.$$ Since $n \ne 0,$ we get $n^2 - 1 = 24,$ so $n^2 = 25$ and $n = 5.$ The three integers are $4,$ $5,$ and $6,$ and $4^2 + 5^2 + 6^2 = 16 + 25 + 36$ $= 77.$ Thus, the correct answer is B.