Show that `n^3+(n+1)^3+(n+2)^3` is divisible by `9` for every natural number `n`.

Explanation

Let f(n)=`n^(3)+(n+1)^(3)+(n+2)^(3)`
Let P(n): `f(n)i.e. n^(3)+(n+1)^(3)+(n+2)^(3)` is divisible by 9.
Now, f(1)=`1^(3)+2^(3)+3^(2)=36` is divisible by 9.
`therefore P(1) ` is true …….(A)
Let P(m) be true `Rightarrow f(m)=m^(3)+(m+1)^(3)+(m+2)^(3)` is divisible by 9.
`Rightarrow f(m)=m^(3)+(m+1)^(3)+(m+2)^(3)=9k` where k is an integer ...(i)
To prove P(m+1) is true i.e. f(m+1) is divisible by 9.
Now, `f(m+1)=m(m+1)^(3)+(m+2)^(3)+(m+3)^(3)`
`=(m+1)^(3)+(m+2)^(3)+m^(3)+9m^(2)+27m+27`
`=[m^(3)+(m+1)^(3)+(m+2)^(3)]+9m^(2)+27m+27`
`=9k+9(m^(2)+3m+3)`
which is divisible by 9
Hence P(m+1) is true whenever P(m) is true. .........(B)
From (A) and (B). It follows that P(n) is true for all natural number n.