Class 8 Maths Chapter 6 Important Question Answers - Cubes and Cube Roots

Q1: Find out the cube root of 13824 by the prime factorisation method.
Sol: First, let us prime factorise 13824:
13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
= 2 ³ × 2 ³ × 2 ³ × 3 ³
3√13824 = 2 × 2 × 2 × 3 = 24

Q2: (13/10) ³ 
Sol: The cube of a rational number is the result of multiplying a number by itself three times.
To evaluate the cube of (13/10) ³
Firstly we need to convert into proper fractions, i.e.(13/10) ³
We need to multiply the given number three times, i.e. (13/10) × (13/10) × (13/10) = (2197/1000)
∴ the cube of (1 3/10) is (2197/1000)

Q3: Find the cube root of 10648 by the prime factorisation method.
Sol:10648 = 2 × 2 × 2 × 11 × 11 × 11
Grouping the factors in triplets of number equal factors,
10648 = (2 × 2 × 2) × (11 × 11 × 11)
Here, 10648 can be grouped into triplets of number equal factors,
∴ 10648 = 2 × 11 = 22
Therefore, the cube root of 10648 is 22.

Q4: The cube of 100 will have _________ zeroes.
Sol: The cube of 100 will have  six zeroes.
= 1003
= 100 × 100 × 100
= 1000000

Q5: (1.2) ³ = _________.
Sol: (1.2) ³  = 12/10
= (12/10) × (12/10) × (12/10)
= 1728/1000
= 1.728

Q6: Find the cube root of 91125 by the prime factorisation method.
Sol: 91125 = 3 × 3 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5
By grouping the factors in triplets of equal factors, 91125 = (3 × 3 × 3) × (3 × 3 × 3) × (5 × 5 × 5)
Here, 91125 can be grouped into triplets of equal factors,
∴ 91125 = (3 × 3 × 5) = 45
Thus , 45 is the cube root of 91125.

Q7: A cuboid of plasticine made by Parikshit with sides 5 cm, 2 cm, and 5 cm. How many such cuboids will be needed to form a cube?
Sol: The given side of the cube is 5 cm, 2 cm and 5 cm.
Therefore, volume of cube = 5 × 2 × 5 = 50
The prime factorisation of 50 = 2 × 5 × 5
Here, 2, 5 and 5 cannot be grouped into triples of equal factors.
Therefore, we will multiply 50 by 2 × 2 × 5 = 20 to get the perfect square.
Hence, 20 cuboids are needed to form a cube.

Q8: State true or false.
(i) The cube of any odd number is even
(ii) A perfect cube never ends with two zeros.
(iii) If the 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.
Sol:
(i) This statement is false.
Taking a cube of any required odd numbers
3³= 3 x 3 x 3 = 27
7³=7 x 7 x 7= 343
5³=5 x 5 x 5=125
All the required cubes of any given odd number will always be odd.
(ii) This statement is true.
10³= 10 x 10 x 10= 1000
20³ = 20 x 20 x 20 = 2000
150³ =150 x150 x150 = 3375000
Hence a perfect cube will never end with two zeros.
(iii) This statement is false.
15²= 15 x15= 225
15³= 15 x 15 x 15= 3375
Thus, the square of any given number ends with 5; then the cube ends with the number 25 is an incorrect statement.
(iv) This statement is false.
2³= 2x2x2= 8
12³ = 12 x 12 x 12= 1728
Accordingly, There are perfect cubes ending with the number 8
(v) This statement is false.
The minimum two digits number is 10
And 
10³=1000→4 Digit number.
The maximum two digits number is 99
And 
99³=970299→6 Digit number
Accordingly, the cube of two-digit numbers can never be a three-digit number.
(vi) This statement is false 
10³=1000→4 Digit number.
The maximum two digits number is 99
And 
99³=970299→6 Digit number
Accordingly, the cube of two-digit numbers can never have seven or more digits.
(vii) This statement is true
1³ = 1 x 1 x 1= 1
2³ = 2 x 2 x 2= 8
According to the cube, a single-digit can be a single-digit number.

Q9: Find the cube of 3.5.
Sol: 3.53 = 3.5 x 3.5 x 3.5
= 12.25 x 3.5
= 42.875

Q10: There are _________ perfect cubes between 1 and 1000.
Sol: 
There are 8 perfect cubes between 1 and 1000.
2 × 2 × 2 = 8
3 × 3 × 3 = 27
4 × 4 × 4 = 64
5 × 5 × 5 = 125
6 × 6 × 6 = 216
7 × 7 × 7 = 343
8 × 8 × 8 = 512
9 × 9 × 9 = 729

Q11: Is 392 a perfect cube? If not, find the smallest natural number by which 392 should be multiplied so that the product is a perfect cube.
Sol: The prime factorisation of 392 gives:
392 = 2 x 2 x 2 x 7 x 7
As we can see, number 7 cannot be paired in a group of three. Therefore, 392 is not a perfect cube.
We must multiply the 7 by the original number to make it a perfect cube.
Thus,
2 x 2 x 2 x 7 x 7 x 7 = 2744, which is a perfect cube, such as 23 x 73 or 143.
Hence, the smallest natural number, which should be multiplied by 392 to make a perfect cube, is 7.

Q12: Which of the following numbers are in perfect cubes? In the case of a perfect cube, find the number whose cube is the given number 256 
Sol: A perfect cube can be expressed as a product of three numbers of equal factors
Resolving the given number into prime factors, we obtain
256 = 2 × 2 × 2 × 2 × 2× 2 × 2 × 2
Since the number 256 has more than three factors
∴ 256 is not a perfect cube.

Q13: Find the smallest number by which 128 must be divided to get a perfect cube.
Sol:  The prime factorisation of 128 is given by:
128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
By grouping the factors in triplets of equal factors,
128 = (2 × 2 × 2) × (2 × 2 × 2) × 2
Here, 2 cannot be grouped into triples of equal factors.
Therefore, to obtain a perfect cube, we will divide 128 by 2.