Prove the statement by the Principle of Mathematical Induction: $n^{3}-7n+3$ is divisible by $3$ for all natural numbers $n$.
(N/A) Let $P(n): n^{3}-7n+3$ be divisible by $3$ for all $n \in N$.
Step $1$: For $n=1$,$P(1) = (1)^{3}-7(1)+3 = 1-7+3 = -3$.
Since $-3$ is divisible by $3$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N$,i.e.,$k^{3}-7k+3 = 3m$ for some integer $m$.
Step $3$: We need to show $P(k+1)$ is true.
$P(k+1) = (k+1)^{3}-7(k+1)+3$
$= (k^{3}+3k^{2}+3k+1) - 7k - 7 + 3$
$= (k^{3}-7k+3) + 3k^{2}+3k-6$
$= 3m + 3(k^{2}+k-2)$
$= 3(m+k^{2}+k-2)$.
Since $3(m+k^{2}+k-2)$ is divisible by $3$,$P(k+1)$ is true.
Thus,by the Principle of Mathematical Induction,$P(n)$ is true for all $n \in N$.
Step $1$: For $n=1$,$P(1) = (1)^{3}-7(1)+3 = 1-7+3 = -3$.
Since $-3$ is divisible by $3$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N$,i.e.,$k^{3}-7k+3 = 3m$ for some integer $m$.
Step $3$: We need to show $P(k+1)$ is true.
$P(k+1) = (k+1)^{3}-7(k+1)+3$
$= (k^{3}+3k^{2}+3k+1) - 7k - 7 + 3$
$= (k^{3}-7k+3) + 3k^{2}+3k-6$
$= 3m + 3(k^{2}+k-2)$
$= 3(m+k^{2}+k-2)$.
Since $3(m+k^{2}+k-2)$ is divisible by $3$,$P(k+1)$ is true.
Thus,by the Principle of Mathematical Induction,$P(n)$ is true for all $n \in N$.
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