Answer 1

\[\text{ Let } p\left( n \right) = n^3 - 7n + 3 \text{ is divisible by } 3 \forall n \in N . \]

\[\text{ Step I: For }  n = 1, \]

\[p\left( 1 \right) = 1^3 - 7 \times 1 + 3 = 1 - 7 + 3 = - 3, \text{ which is clearly divisible by } 3\]

\[\text{ So, it is true for n } = 1\]

\[\text{ Step II: For }  n = k, \]

\[\text{ Let } p\left( k \right) = k^3 - 7k + 3 = 3m, \text{ where m is any integer, be true } \forall   k \in N . \]

\[\text{ Step III: For }  n = k + 1, \]

\[p\left( k + 1 \right) = \left( k + 1 \right)^3 - 7\left( k + 1 \right) + 3\]

\[ = k^3 + 3 k^2 + 3k + 1 - 7k - 7 + 3\]

\[ = k^3 + 3 k^2 - 4k - 3\]

\[ = k^3 - 7k + 3 + 3 k^2 + 3k - 6\]

\[ = 3m + 3\left( k^2 + k + 2 \right) \left[ \text{ Using step }  II \right]\]

\[ = 3\left( m + k^2 + k + 2 \right)\]

\[ = 3p, \text{ where p is any integer } \]

\[\text{ So,}   p\left( k + 1 \right) \text{ is divisible by } 3 .\]

Hence, n³  - 7n + 3 is divisible by 3 for all n \[\in\] N .