**1. ** (i) 216

Prime factors of 216 = 2 × 2 × 2 × 3 × 3 × 3

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

Therefore, 216 is a perfect cube number.

(ii) 128

Prime factors of 216 = 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 216 = 2 × 2 × 2 × 3 × 3 × 3

Here all factors appear in 3’s group

Therefore, 1000 s a perfect cube.

(iv) 100

Prime factors of 216 = 2 × 2 × 5 × 5

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

Therefore, 100 is not a perfect cube.

(v) 46656

Prime factors of 216 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

Here all factors appear in 3’s group

Therefore, 46656 s a perfect cube.

**2.** (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.

**3. ** (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 factor 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.

**4. ** Given numbers = 5 × 2 × 5

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

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

Hence he needs 20 cuboids.

**5. ** (i) 64

(ii) 512

(iii) 10648

(iii) 27000

(v) 15625

(vi) 13824

(vii) 110592

(viii) 175616

(ix) 91125

**6. ** (i) False

Since, 1^{3} = 1, 3^{3} = 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, ………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 [Ends with 8]

And 22^{3 }= 10648 [Ends with 8]

(v) False

Since 10^{3 }= 1728 [Four digit number]

And 11^{3 }= 10648 [Four digit number]

(vi) False

Since 99^{3} = 970299 [Six digit number]

(vii) True

1^{3} = 1 [Six digit number]

2^{3} = 8 [Single digit number]

**7. ** 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^{5} = 27

Here in cube, ones digit is 7

Now next number with 3 as ones digit 13^{5} = 2197

And next number with 3 as ones digit 23^{5} = 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.