Q. 225.0( 2 Votes )

# 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.

(ii) If n leaves 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 then n^{3} also a number of the same type.

Answer :

(i) If n is even, then n^{3} is also even.

Let the three even natural numbers be 2 , 4 , 6

Cubes of these numbers ,

= 2^{3} = 8

= 4^{3} = 64

= 6^{3} = 216

Hence, we can see that all cubes are even in nature.

Statement verified.

(ii) If n is odd, then n^{3} is also odd.

Let three odd natural numbers are = 3 , 5 , 7

Cubes of these numbers =

= 3^{3} = 27

= 5^{3} = 125

= 7^{3} = 343

Hence, we can see that all cubes are odd in nature.

Statement verified.

Let three natural numbers of the form (3n+1) are = 4, 7 , 10

Cube of numbers = 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.

Statement verified.

(iv) If a natural number n is of the form 3p+2 then n^{3} also a number of the same type.

Let three natural numbers of the form (3p+2) are = 5 , 8 , 11

Cube of these numbers = 5^{3} = 125 , 8^{3} = 512 , 11^{3} = 1331

Now, we try to write these cubes in form of (3p + 2)

= 125 = 3 × 41 + 2

= 512 = 3 × 170 + 2

= 1331 = 3 × 443 + 2

Hence, statement verified.

Rate this question :

Which of the following numbers are not perfect cubes?

(i) 64

(ii) 216

(iii) 243

(iv) 1728

RD Sharma - MathematicsFind the cubes of the following numbers by column method:

(i) 35

(ii) 56

(iii) 72

RD Sharma - Mathematics