By taking three different values of n verify the truth of the following statements:
(i) If n is even , then n3 is also even.
(ii) If n is odd, then n3 is also odd.
(iii) If n leaves remainder 1 when divided by 3, then n3 also leaves 1 as remainder when divided by 3.
(iv) If a natural number n is of the form 3p + 2 then n3 also a number of the same type.
(i)
Let the three even natural numbers be 2, 4 and 8.
Cubes of these numbers are:
$2^{3}=8,4^{3}=64,8^{3}=512$
By divisibility test, it is evident that 8,64 and 512 are divisible by 2 .
Thus, they are even.
This verifies the statement.
(ii)
Let the three odd natural numbers be 3, 9 and 27.
Cubes of these numbers are:
$3^{3}=27,9^{3}=729,27^{3}=19683$
By divisibility test, it is evident that 27,729 and 19683 are divisible by $3 .$
Thus, they are odd.
This verifies the statement.
(iii)
Three natural numbers of the form $(3 n+1)$ can be written by choosing $n=1,2,3 \ldots$ etc.
Let three such numbers be 4,7 and 10 .
Cubes of the three chosen numbers are:
$4^{3}=64,7^{3}=343$ and $10^{3}=1000$
Cubes of 4,7 and 10 can expressed as:
$64=3 \times 21+1$, which is of the form $(3 n+1)$ for $n=21$
$343=3 \times 114+1$, which is of the form $(3 n+1)$ for $n=114$
$1000=3 \times 333+1$, which is of the form $(3 n+1)$ for $n=333$
Cubes of 4,7, and 10 can be expressed as the natural numbers of the form $(3 n+1)$ for some natural number $n$. Hence, the statement is verified.
(iv)
Three natural numbers of the form $(3 p+2)$ can be written by choosing $p=1,2,3 \ldots$ etc.
Let three such numbers be 5,8 and 11 .
Cubes of the three chosen numbers are:
$5^{3}=125,8^{3}=512$ and $11^{3}=1331$
Cubes of 5,8, and 11 can be expressed as:
$125=3 \times 41+2$, which is of the form $(3 p+2)$ for $p=41$
$512=3 \times 170+2$, which is of the form $(3 p+2)$ for $p=170$
$1331=3 \times 443+2$, which is of the form $(3 p+2)$ for $p=443$
Cubes of 5,8, and 11 could be expressed as the natural numbers of the form $(3 p+2)$ for some natural number $p$. Hence, the statement is verified.