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

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Let P(n) : `n^(3)-7n+3` is divisible by 3, for all natural number n.
Step I We observe that P(1) is true.
`P(1)=(1)^(3)-7(1)+3`
=11-7+3
-3, which is divisible by 3.
Hence, P(1) is true. Step II Now, assume that P(n) is true for n=k.
`P(k+1):(k+1)^(3)-7(k+1)+3`
`=k^(3)+1+3k(k+1)-7k-7+3`
`=k^(3)-7k+3+3k(k+1)-6`
`=3q+3[k(k+1)-2]`
Hence, P(k+1) is true whenever P(k) is true. [from stem II]
So, by the principle of mathematical induction P(n): is true for all natural number n.
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