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.

(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.

(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.

(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.

**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.

(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.

(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

(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.

(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.

**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.

(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.

(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.

(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.

**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) cm^{3}

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

(ii) 512

^{3}√512 = √(2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2)

^{3}√512 = 2 * 2 * 2

^{3}√512 = 8

(iii) 10648

^{3}√10648 = √(2 * 2 * 2 * 11 * 11 * 11)

^{3}√10648 = 2 * 11

^{3}√10648 = 22

(iv) 27000

^{3}√27000 = √(2 * 2 * 2 * 3 * 3 * 3 * 5 * 5 * 5)

^{3}√27000 = 2 * 3 * 5

^{3}√27000 = 30

(v) 15625

^{3}√15625 = √(5 * 5 * 5 * 5 * 5 * 5)

^{3}√15625 = 5 * 5

^{3}√15625 = 25

(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

(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

(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

(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

(x) 91125

^{3}√91125 = √(3 * 3 * 3 * 3 * 3 * 3 * 5 * 5 * 5)

^{3}√91125 = 3 * 3 * 5

^{3}√91125 = 45

**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 1^{3} = 1, 3^{3} = 27, 5^{3} = 125, ………….. are all odd.

(ii) True

A perfect cube ends with three zeroes.

Ex: 10^{3} = 1000, 20^{3} = 8000, 30^{3} = 27000, ………….. so on.

(iii) False.

Since 5^{2} = 25, 5^{3} = 125, 15^{2} = 225, 15^{3} = 3375 (Did not end with 25)

(iv) False

Since 12^{3} = 1728 (End with 8)

And 22^{3} = 10648 (End with 8)

(v) False

Since 10^{3} = 1000 (Four digit number)

And 11^{3} = 1331 (Four digit number)

(vi) False

Since 99^{3} = 970299 (Six digit number)

(vii) True

1^{3} = 1 (Single digit number)

2^{3} = 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

10^{3 }= 1000 and Possible cube of

11^{3} = 1331

Since, cube of unit’s digit

1^{3} = 1

Therefore, cube root of 1331 is 11.

4913

We know that

7^{3} = 343

Next number comes with 7 as unit place

17^{3} = 4913

Hence, cube root of 4913 is 17.

12167

We know that

3^{3} = 27

Here in cube, ones digit is 7

Now next number with 3 as ones digit

13^{3} = 2197

And next number with 3 as ones digit

23^{3} = 12167

Hence cube root of 12167 is 23.

32768

We know that

2^{3} = 8

Here in cube, ones digit is 8

Now next number with 2 as ones digit

12^{3} = 1728

And next number with 2 as ones digit

22^{3} = 10648

And next number with 2 as ones digit

32^{3} = 32768

Hence cube root of 32768 is 32.

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