Prove the following by the principle of mathematical induction: `n^3-7n+3` is divisible 3 for all `n in Ndot`

Explanation

Let `p(n)=n^{3}-7 n+3` is divisible by `3 forall n in N`.

Step I: For `n=1`,

`p(1)=1^{3}-7 times 1+3=1-7+3=-3`, which is claarly divisible by 3

So, it is true for `n=1`

Step ll: For `n=k`,

Let `p(k)=k^{3}-7 k+3=3 m`, where `m` is any integer, be true `forall k in N`.

Step III: For `n=k+1`,

`p(k+1)=(k+1)^{3}-7(k+1)+3`

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