The document NCERT Solutions - Chapter 7 : Cubes and Cubes Roots, Maths, Class 8 | EduRev Notes is a part of Class 8 category.

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**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 = 2x2x2x3x3x3

Here all factors are in groups of 3’s [in triplets) Therefore, 216 is a perfect cube number.

(ii)

Prime factors of 128 =2x2x2x2x2x2x2

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

Therefore, 128 is not a perfect cube.

(iii)

Prime factors of 1000 =2x2x2x3x3x3

Here all factors appear in 3's group.

Therefore, 1000 is a perfect cube.

(iv)

Prime factors of 100 =2x2x5x5

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

Therefore, 100 is not a perfect cube.

(v)

Prime factors of 46656 = 2x2x2 x2x2x2x3x3x3x3x3x3

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 2: **

**(i) **

Prime factors of 243 =3x3x3x3x3

Here 3 does not appear in 3's group.

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

(ii)

Prime factors of 256 = 2 x 2 x 2 x 2x2x 2x2x2

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

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

(iii)

Prime factors of 72 = 2x2x2x3x3

Here 3 does not appear in 3's group.

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

(iv)

Prime factors of 675 =3x3x3x5x5

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

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

(v)

Prime factors of l00 = 2x2x5x5

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

Therefore 100 must be multiplied by 2 x 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 3:**

(i)

Prime factors of 81 = 3x3x3x3

Here one factor 3 is not grouped in triplets.

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

(ii)

Prime factors of 128 =2x2x2x2x2x2x2

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)

Prime factors of 135 = 3x3x3x5

Here one factor 5 does not appear in a triplet.

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

(iv)

Prime factors of 192 =2x2x2x2x2x2x3

Here one factor 3 does not appear in a triplet

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

(v)

Prime factors of 704 = 2x2x2x2x2x2x11

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 4: **

Given numbers = 5 x 2 x 5

Since, Factors of 5 and 2 both are not in group of three.

Therefore, the number must be multiplied by 2 x 2 x 5 = 20 to make it a perfect cube.

Hence he needs 20 cuboids.

**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 1: **

**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^{2 }= 27,5^{3} = 125,. ......... are all odd.

(ii) True

Since, a perfect cube ends with three zeroes.

e.g. 10^{3} = 1000,20^{3} =8000,30^{3} = 27000,...... soon

(iii) False

Since, *5 ^{2}* = 25,5

(iv) False

Since 12^{3 }=1728 [Ends with 8]

And 22^{3 }=10648 [Ends 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 I^{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

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 I2^{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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