By taking three different values of n verify the truth of the following statements: (i) If `n` is even, then `n^3` is also even.
(ii) if `n` is odd, then `n^3` is also odd.
(iii) If `n` leaven remainder 1 when divided by 3, then `n^3` also leaves 1 as remainder when divided by 3.
(iv) If a natural number n is of the form `3p+2\ t h e n\ n^3` also a number of the same type.

Explanation

(i) If `n` is even, then `n^3` is also even.
Let three even natural numbers `2`, `4`, `6`.
So now,
`2^3=8`
`4^3=64`
`6^3=216`
Hence, we can see that all cubes are even in nature.
Thus, the given Statement is verified.

(ii) If n is odd, then `n^3` is also odd.
Let three odd natural numbers `3`, `5`, `7`.
So now,
`3^3=27`
`5^3=125`
`7^3=343`
Hence, we can see that all cubes are odd in nature.
Thus, the given statement is verified.

(iii) If `n` leaves remainder `1` when divided by `3`, then `n^3` also leaves `1` as remainder when divided by `3`.
Let three natural numbers of the form `(3n+1)` are `4`, `7` and `10`.
So now,
`4^3=64`
`7^3=343`
`10^3=1000`
We can see that if we divide these numbers by `3`, we get `1` as remainder in each case.
Hence, statement is verified.

(iv) If a natural number `n` is of the form `(3p+2)` then `n^3` also a number of the same type.
Let us consider three natural numbers of the form `(3p+2) `are `5`, `8` and `11`.
So now,
`5^3=125`
`8^3=512`
`11^3=1331`
Now, we try to write these cubes in form of `(3p + 2)`
`125=3xx41+2`
`512=3xx170+2`
`1331=3xx443 +2`
Hence, the given statement is verified.