Exercise 7.1
Question 1.
Which of the following numbers are not perfect cubes:
(i)216
(ii)128
(iii)1000
(iv)100
(v)46656
Solution :
(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 x 2 x 5 x 5
Here all factors do not appear in 3’s group.
Therefore, 100 is not a perfect cube.
(v) 46656
Prime factors of 46656
= 2x2x2x2x2x2x3x3x3x3x3x3
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
Solution :
(i)243
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) 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 
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
Solution :
(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 =
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 =
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 =
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
=
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?
Ans. Given numbers =
Since, Factors of 5 and 2 both are not in group of three.
Therefore, the number must be multiplied by 2 × 5 × 2 = 20 to make it a perfect cube.
Hence he needs 20 cuboids.
NCERT Solutions for Class 8 Maths 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
Solution : (i) 64
= 4
(ii) 512
(iii) 10648
(iv) 27000
(v) 15625
=5 × 5
= 25
(vi) 13824
=
= 24
(vii) 110592
= 2×2×2×2×3
= 48
(viii) 46656
= 36
(ix) 175616
= 56
(x) 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.
Solution : (i) False
Since,
are all odd.
(ii) True
Since, a perfect cube ends with three zeroes. e.g.
so on
(iii) False
Since,
(Did not end with 25)
(iv) False
Since 
= 1728
[Ends with 8]
And = 10648
[Ends with 8]
(v) False Since 
= 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.
Solution :
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.