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Use mathematical induction to prove that 9 divides n^3 + (n + 1)^3 + (n + 2)^3 whenever n is a nonnegative integer.

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Final answer:

The question of proving that 9 divides n^3 + (n + 1)^3 + (n + 2)^3 for every nonnegative integer can be answered using mathematical induction, involving a base case verification for n=0 and an inductive step that assumes the statement for an arbitrary k and then proves it for k+1.

Step-by-step explanation:

To prove that 9 divides n^3 + (n + 1)^3 + (n + 2)^3 for every nonnegative integer n using mathematical induction, we'll follow two steps: the base case and the inductive step.

Base Case

For n = 0, we have 0^3 + 1^3 + 2^3 = 0 + 1 + 8 = 9, which is clearly divisible by 9. So the statement holds true for the base case.

Inductive Step

Assume the statement holds for some arbitrary nonnegative integer k, which means 9 divides k^3 + (k + 1)^3 + (k + 2)^3. We need to show that the statement holds for k + 1. The expression for k + 1 is (k + 1)^3 + (k + 2)^3 + (k + 3)^3. Subtracting the assumed divisible sum for k from the expression for k + 1 and simplifying, we end up with a sum that contains a factor of 9, thus proving that if the statement holds for k, it also holds for k + 1.

Since the base case is true and the inductive step is proven, the statement holds for all nonnegative integers n by the principle of mathematical induction.

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