Karnataka Class 8 Math Solution Chapter Cubes and Cube Roots by Experience Math Teacher. Karnataka Board Class 8 Math Part 2 PDF Chapter 1 Exercise 1.1, Exercise 1.2 Solution.

**Karnataka Class 8 Math Solution Chapter Cubes and Cube Roots**

Cubes and Cube Roots Karnataka Class 8 PDF**Part 2 Chapter 1 Exercise 1.1**

**(1) Which of the following number are not perfect cubs?**

**(i) 216**

=> Solution:- By prime factorization

216 = 2 x 2 x 2 x 3 x 3 x 3

Each factor appears 3 time

216 = 2^{3} x 3^{3}

= (2 x 3)^{3}

= 63 which is perfect cube.

So 216 is a perfect cube of 6.

**(ii) 128 **

=> Solution:- by prime factorization

128 = 2 x 2 x 2 x 2 x 2 x 2 x 2

In above factorization one triplet group of 2 is not form.

So, 128 is not a perfect cube.

**(iii) 1000**

=> Solution:- By prime factorization

100 = 5 x 5 x 5 x 2x 2 x 2

Here each factor appear three time

1000 = 5^{3} x 2^{3}

= (5 x 2)^{3}

= 10^{3 }which is a perfect cube

So, 1000 is a perfect cube of 10.

**(iv) 100**

=> Solution:- By prime factorization

100 = 5 x 5 x 2 x 2

Here each factor repeated only twice.

So, 100 is not a perfect cube.

**(v) 46656**

=> Solution: – By prime factorization

46656 = 3 x 3 x 3 x 3 x 3 x 3 x 4 x 4 x 4

Here each factor appear 3 time

46656 = 3^{3} x 3^{3} x 4^{3}

= (3 x 3 x 4)^{4}

= (36)^{3} which is a perfect cube.

So, 46656 is a perfect cube of 36.

**(2) Find the smallest number by which each of the following number must be multiplied to obtain a perfect cube.**

(i) Solution:- By smallest multiple method

243 = 3 x 3 x 3 x 3 x 3

Here, the prime factor 3 does not appear in a group of three, to obtain a perfect cube, we need one more 3.

In that case

243 x 3 = 3 x 3 x 3 x 3 x 3 x 3

= 729 which is a perfect cube.

Hence the smallest natural number by which 243 should be multiplied to make a perfect cube is 3.

(ii) 256

=> Solution:

Here,

The prime factor 4 does not appear in a group of there. To obtain a perfect cube we need only 2 to make a prime factor 4 i.e. (2 x 2) = 4

In that case

256 x 2 = 2 x 2 x 2 x 4 x 4 x 2 x 2

= 2 x 2 x 2 x 4 x 4x 4 x 2 x 2

= 2 x 2 x 2 x 4 x 4 x 4

= 512 which is a perfect cube.

Hence the smallest natural number by which 256 should be multiplied to make 9 perfect cube is 2.

(iii) 72

=> Solution:-72 = 2 x 2 x 2x 3 x 3

Here,

The prime factor 3 does not appear in a group of three. To obtain a perfect cube we need 3

So, 72 x 3 = 2 x 2 x 2 x 3 x 3 x 3

= 216 which is a perfect cube.

Hence the smallest natural number by which 72 should be multiplied to make a perfect cube is 3.

(iv) 675

=> Solution: 675 = 3 x 3 x 3 x 5 x 5

Here,

The prime factor 5 does not appear in a group of three. To obtain a perfect cube we need one more 5.

In that case

675 x 5 = 3 x 3 x 3 x 5 x 5 x 5

= 3375 which is a perfect cube.

Hence the smallest natural number by which 675 should be multiplied to make a perfect cube is 5.

(v) 100

=> Solution:- 100 = 5 x 5 x 2 x 2

Here both the prime factor does not form a group of three. To obtain a perfect cube we need 5 and 2 i.e. 5 x 2 = 10

Then, 100 x 5 x 2 = 5 x 5 x 5 x 2 x 2 x 2

= 1000 which is a perfect cube.

Hence the smallest natural number by which 100 should be multiplied to make a perfect cube is 5 and 2 or 10.

**(3) Find the smallest number by which each of the following number must be divided to obtain a perfect cube.**

(i) 81

=> Solution:- 81 = 3 x 3 x 3 x 3

In the factorization of 81 remaining prime factor 3 does not form a group of three. So, if we divide 81 by 3, then the prime factorization of the quotient will not contain 3.

So, 81 ÷ 3 = 3 x 3 x 3

= 27

Hence the smallest number by which 81 should be divided to make it perfect cube is 3.

∴ The perfect cube in that case is = 27

(ii) 128

=> Solution:-

128 = 2 x 2 x 2 x 2 x 2 x 2 x 2

In the factorization of 128 prime factor 2 does not form a group of three, so, if we divide 128 by 2, then the prime factorization of te quotients will not contain 2.

So, 128 ÷ 2 = 2 x 2 x 2 x 2 x 2 x 2

= 64

Hence the smallest number by which 128 should be divided to make it perfect cube is 2.

The perfect cube in this case is 64.

(iii) 135

=> Solution:- 135 = 3 x 3 x 3 x 5

In the factorization of 135 the prime factor 5 does not form a group of three so, if we divide 135 by 5, then the prime factorization of the quotient will not contain 5.

So, 135 ÷ 5 = 3 x 3 x 3

= 27

Hence the smallest number by which 135 should be divided to make it perfect cube is 5.

(iv) 192

=> Solution:- 192 = 4 x 4 x 4 x 3

In the factorization of 192 the prime factor 3 does not form a group of there so, if we divide 192 by 3 then the prime factorization of the quotients will not contain 3.

So, 192 ÷ 3 = 4 x 4 x 4

= 64

Hence the smallest number by which 192 should be divided to make it perfect cube is 3.

(v) 704

Solution:- 704 = 2 x 2 x 2 x 2 x 2 x 2 x 11

In the factorization of 704 the prime factor 11 does not form a group of three so, if we divide 704 by 11 then the prime factorization of the quotient will not contain 11.

704 ÷ 11 = 2 x 2 x 2 x 2 x 2 x 2

= 64

Hence the smallest number by which 704 should be divided to make it perfect cube is 11.

**(4) Parikshit makes a cuboid of plasticine of sides 5cm, 2cm, 5cm, how many such cuboids will he need to form a cube?**

=> Solution: Parikshit have plasticine of sides 5cm, 2cm and 5cm

Volume of cuboid is 5 x 2 x 5

= 5 x 2 x 5

= 5 x 5 x 2

Since there is only one 2 in the prime factorization so, we need 2 x 2 i.e. 4 also we need one more 5 to make a group of 5 so we can make a perfect cube

Therefore, we need 20 i.e. (5 x 4) such cuboid to make a cube.

**Part 2 Chapter 1 Exercise 1.2**

**(1) Find the cube root of each of the following numbers by prime factories method.**

(i) 64

=> Solution:- By prime factor method 64 = 2 x 2 x 2 x 2 x 2 x 2

= 2^{3} x 2^{3}

= (2 x 2)^{3}

3√64 = 2 x 2 = 4

Therefore, the cube root of 64 is 4

(ii) 512

=> Solution:- By prime factor method

512 = 2 x 2 x 2 x 4 x 4 x 4

= 2^{3} x 4^{3 }

= (2 x 4)^{3}

3√512 = 8

Therefore, the cube root of 512 is 8.

(iii) 10648

=> Solution:-

By prime factor method

10648

=> Solution:- By prime factor method

10648 = 2 x 2 x 2 x 11 x 11 x 11

= 23 x 11^{3}

= (2 x 11)^{3}

3√10648 = 2 x 11 = 22

The cube root of 10648 is 22

(iv) 27000

=> Solution:- By prime factor method

27000 = 3 x 3 x 3 x 10 x 10 x 10

= 33 x 103

= (3 x 10)3

3v27000 = 3 x 10 = 30

The cube root of 27000 is 30.

(v) 15625

=> Solution:- By prime factor method

15625 = 5 x 5 x 5 x 5 x 5 x 5

= 53 x 53

= (5 x 5)3

3√15625 = 25

The cube root of 15625 is 25.

(vi) 13824

=> Solution:- By prime factor method

13824 = 2 x 2 x 2 x 3 x 3 x 3x 4 x 4 x 4

= 23 x 33 x 43

= (2 x 3 x 4)3

3√13824 = 2 x 3 x 4 = 24

The cube root of 13824 is 24.

(vii) 110592

=> Solution:- By prime factorization method

110592 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 6 x 6 x 6

= 2^{3} x 2^{3} x 2^{3} x 6^{3}

= ( 2 x 2 x 2 6)^{3}

3√110592 = 48

The cube root of 110592 is 48.

(viii) 46656

=> Solution:- By prime factorization method

46656 = 3 x 3 x 3 x 3 x 3 x 3 x 4 x 4 x 4

= 3^{3} x 3^{3} x 4^{3}

= (3 x 3 x 4)^{3}

3√46656 = 36

The cube root of 46656 is 36.

__(2) State true or false__

(i) Cube of any odd number is even

Solution: False.

(ii) A perfect cube does not end with two zero.

Solution: True.

(iii) If square of a number end with 5, then its cube end with 25.

Solution: False.

(iv) There is no perfect cube which ends with 8.

Solution: False.

(v) The cube of a two digit number may be three digit numbers.

Answer: False.

(vi) The cube of a two digit number may have seven or more digits.

Solution: False.

(vii) The cube of a single digit number may be a single digit number.

Solution: True.

**(3) You are told that 1,332 is a perfect cube. Can you guess without factorization what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768**

=> Solution:- 1331 is a perfect cube of 11.

Similarly,

4913 is a perfect cube of 17

12161 is a perfect cube of 23

32768 is a perfect cube of 32