Class 8 - Maths - Cubes and Cube Roots

Exercise 7.1

Question 1:

Which of the following numbers are not perfect cubes:

(i) 216                  (ii) 128             (iii) 1000                (iv)100              (v) 46656

Answer:

(i) 216

Prime factors of 216 = 2 * 2 * 2 * 3 * 3 * 3

Here all factors are in groups of 3’s (in triplets)

Therefore, 216 is a perfect cube number.

Class_8_Cube_&_Cube_Roots_Of_216

(ii) 128

Prime factors of 128 = 2 * 2 * 2 * 2 * 2 * 2 * 2

Here one factor 2 does not appear in a 3’s group.

Therefore, 128 is not a perfect cube.

 

 

(iii) 1000

Prime factors of 1000 = 2 * 2 * 2 * 3 * 3 * 3

Here all factors appear in 3’s group.

Therefore, 1000 is a perfect cube.

 Class_8_Cube_&_Cube_Roots_Of_1000

 (iv)100

Prime factors of 100 = 2 * 2 * 5 * 5

Here all factors do not appear in 3’s group.

Therefore, 100 is not a perfect cube.

Class_8_Cube_&_Cube_Roots_Of_100

(v) 46656

Prime factors of 46656 = 2 * 2 * 2 * 2 * 2 * 2 * 3 * 3 * 3 * 3 * 3 * 3

Here all factors appear in 3’s group.

Therefore, 46656 is a perfect cube.

 Class_8_Cube_&_Cube_Roots_Of_46656

 

Question 2:

Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube:

(i) 243              (ii) 256               (iii) 72              (iv) 675              (v) 100

Answer:

(i) 243

Prime factors of 243 = 3 * 3 * 3 * 3 * 3

Here, 3 does not appear in 3’s group.

Therefore, 243 must be multiplied by 3 to make it a perfect cube.

 Class_8_Cube_&_Cube_Roots_Of_243

(ii) 256

Prime factors of 256 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2

Here one factor 2 is required to make a 3’s group.

Therefore, 256 must be multiplied by 2 to make it a perfect cube.

Class_8_Cube_&_Cube_Roots_Of_256 

 

 (iii) 72

Prime factors of 72 = 2 * 2 * 2 * 3 * 3

Here 3 does not appear in 3’s group.

Therefore, 72 must be multiplied by 3 to make it a perfect cube

Class_8_Cube_&_Cube_Roots_Of_72

(iv) 675

Prime factors of 675 = 3 * 3 * 3 * 5 * 5

Here factor 5 does not appear in 3’s group.

Therefore 675 must be multiplied by 3 to make it a perfect cube.

 Class_8_Cube_&_Cube_Roots_Of_675

(v) 100

Prime factors of 100 = 2 * 2 * 5 * 5

Here factor 2 and 5 both do not appear in 3’s group.

Therefore 100 must be multiplied by 2 * 5 = 10 to

make it a perfect cube.

 Class_8_Cube_&_Cube_Roots_Of_100

 

Question 3:

Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube:

(i) 81             (ii) 128              (iii) 135             (iv) 192                (v) 704

Answer:

(i) 81

Prime factors of 81 = 3 * 3 * 3 * 3

Here one factor 3 is not grouped in triplets.

Therefore 81 must be divided by 3 to make it a perfect cube.

 Class_8_Cube_&_Cube_Roots_Of_81

(ii) 128

Prime factors of 128 = 2 * 2 * 2 * 2 * 2 * 2 * 2

Here one factor 2 does not appear in a 3’s group.

Therefore, 128 must be divided by 2 to make it a perfect cube.

 

(iii) 135

Prime factors of 135 = 3 * 3 * 3 * 5

Here one factor 5 does not appear in a triplet.

Therefore, 135 must be divided by 5 to make it a perfect cube.

 Class_8_Cube_&_Cube_Roots_Of_135

(iv) 192

Prime factors of 192 = 2 * 2 * 2 * 2 * 2 * 2 * 3

Here one factor 3 does not appear in a triplet.

Therefore, 192 must be divided by 3 to make it a perfect cube.

 Class_8_Cube_&_Cube_Roots_Of_192

(v) 704

Prime factors of 704 = 2 * 2 * 2 * 2 * 2 * 2 * 11

Here one factor 11 does not appear in a triplet.

Therefore, 704 must be divided by 11 to make it a perfect cube.

Class_8_Cube_&_Cube_Roots_Of_704 

 

 

Question 4:

Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?

Answer:

Volume of the cube o sides 5 cm, 2 cm 5 cm = (5 * 2 * 5) cm3

Here, two 5s and one 2 are left which are not in a triplet.

If we multiply this expression by 2 * 2 * 5 = 20, then it will be a perfect square.

Thus, 5 * 2 * 5 * 2 * 2 * 5 = 2 * 2 * 2 * 5 * 5 * 5 = 1000

This is a perfect square.

Hence, 20 cuboids of 5 cm, 2 cm, 5 cm are required to form a cube.

  

                                                                 Exercise 7.2

Question 1:

Find the cube root of each of the following numbers by prime factorization method:

(i) 64                       (ii) 512                (iii) 10648              (iv) 27000               (v) 15625           

(vi) 13824             (vii) 110592        (viii) 46656             (ix) 175616             (x) 91125

Answer:

(i) 64      

   3√64 = √(2 * 2 * 2 * 2 * 2 * 2)

   3√64 = 2 * 2

   3√64 = 4

 Class_8_Cube_&_Cube_Roots_Of_64

 (ii) 512               

3√512 = √(2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2)

   3√512 = 2 * 2 * 2

   3√512 = 8

 Class_8_Cube_&_Cube_Roots_Of_512

 

(iii) 10648              

3√10648 = √(2 * 2 * 2 * 11 * 11 * 11)

   3√10648 = 2 * 11 

   3√10648 = 22

 Class_8_Cube_&_Cube_Roots_Of_10648

(iv) 27000                

3√27000 = √(2 * 2 * 2 * 3 * 3 * 3 * 5 * 5 * 5)

   3√27000 = 2 * 3 * 5

   3√27000 = 30

 Class_8_Cube_&_Cube_Roots_Of_27000

 

 

(v) 15625            

3√15625 = √(5 * 5 * 5 * 5 * 5 * 5)

   3√15625 = 5 * 5

   3√15625 = 25

 Class_8_Cube_&_Cube_Roots_Of_15625

(vi) 13824            

3√13824 = √(2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 3 * 3 * 3)

   3√13824 = 2 * 2 * 2 * 3

   3√13824 = 24

 Class_8_Cube_&_Cube_Roots_Of_13824

 

 (vii) 110592        

3√110592 = √(2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 3 * 3 * 3)

   3√110592 = 2 * 2 * 2 * 2 * 3

   3√110592 = 48

 

 Class_8_Cube_&_Cube_Roots_Of_110592

(viii) 46656            

3√46656 = √(2 * 2 * 2 * 2 * 2 * 2 * 3 * 3 * 3 * 3 * 3 * 3)

   3√46656 = 2 * 2 * 3 * 3

   3√46656 = 36

 Class_8_Cube_&_Cube_Roots_Of_46656

 

(ix) 175616             

3√175616 = √(2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 7 * 7 * 7)

   3√175616 = 2 * 2 * 2 * 7

   3√175616 = 56

 

 Class_8_Cube_&_Cube_Roots_Of_110592

 

  

(x) 91125

3√91125 = √(3 * 3 * 3 * 3 * 3 * 3 * 5 * 5 * 5)

   3√91125 = 3 * 3 * 5

   3√91125 = 45

 

 Class_8_Cube_&_Cube_Roots_Of_91125

 

Question 2:

State true or false:

(i) Cube of any odd number is even.

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

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

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

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

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

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

Answer:

(i) False

Since 13 = 1, 33 = 27, 53 = 125, ………….. are all odd.

(ii) True

A perfect cube ends with three zeroes.

Ex: 103 = 1000, 203 = 8000, 303 = 27000, ………….. so on.

(iii) False.

Since 52 = 25, 53 = 125, 152 = 225, 153 = 3375 (Did not end with 25)

(iv) False

Since 123 = 1728 (End with 8)

And 223 = 10648 (End with 8)

(v) False

Since 103 = 1000 (Four digit number)

And 113 = 1331 (Four digit number)

 (vi) False

Since 993 = 970299 (Six digit number)

(vii) True

13 = 1 (Single digit number)

23 = 8 (Single digit number)

Question 3:

You are told that 1,331 is a perfect cube. Can you guess with factorization what is its cube root? Similarly guess the cube roots of 4913, 12167, 32768.

Answer:

We know that

103 = 1000 and Possible cube of

113 = 1331

Since, cube of unit’s digit

13 = 1

Therefore, cube root of 1331 is 11.

 

4913

We know that

73 = 343

Next number comes with 7 as unit place

173 = 4913

Hence, cube root of 4913 is 17.

 

 

12167

We know that

33 = 27

Here in cube, ones digit is 7

Now next number with 3 as ones digit

133 = 2197

And next number with 3 as ones digit

233 = 12167

Hence cube root of 12167 is 23.

 

32768

We know that

23 = 8

Here in cube, ones digit is 8

Now next number with 2 as ones digit

123 = 1728

And next number with 2 as ones digit

223 = 10648

And next number with 2 as ones digit

323 = 32768

Hence cube root of 32768 is 32.

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