RBSE Solutions Class 8 Maths Chapter 2 Cube and Cube Roots Exercise 2.2

Question 1. Determine True/False in the given statements.
(i) Every even number has even cube.
(ii) A perfect cube does not end with double zero (00).
(iii) No one perfect cube end with 8.
(iv) If square of any number is ending with 5 then its cube is end with 25.
(v) Cube of single digit is also a single digits number.
(vi) Cube of double-digit number is of 4 to 6 digits.
Answer:
(i) True. The cube of an even number is always even because multiplying even numbers together always results in an even number.
(ii) True. Perfect cubes can end with three zeros (000) or more, but never exactly two zeros.
(iii) False. For example, \( 2^3 = 8 \) and \( 12^3 = 1728 \), both of which end in 8.
(iv) True. This pattern occurs because numbers ending in 5 have cubes that always end in 5, and specifically, the last two digits will be 25.
(v) False. For instance, \( 3^3 = 27 \) which has two digits, and \( 4^3 = 64 \) also has two digits.
(vi) True. The smallest two-digit number is 10 (\( 10^3 = 1000 \), which has 4 digits) and the largest is 99 (\( 99^3 = 970299 \), which has 6 digits).
In simple words: We check each statement about cubes to see if it is correct or incorrect. Cubing means multiplying a number by itself three times.

🎯 Exam Tip: To quickly check statements about cubes, try a few small examples for both even and odd numbers, or numbers ending in specific digits.

Question 2. Find the cube roots of the following numbers by estimate and prime factor method. Verify your answer
(i) 64
(ii) 343
(iii) 5832
(iv) 74088
(v) 3375
(vi) 10648
(vii) 46656
(viii) 91125
Answer:
(i) 64
**Prime Factorization Method:**
First, we break down 64 into its prime factors:
\( 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \)
Now, we group the factors in sets of three:
\( 64 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \)
\( 64 = 2^3 \times 2^3 \)
\( 64 = (2 \times 2)^3 \)
\( \implies \sqrt[3]{64} = 2 \times 2 = 4 \)
**Estimation Method:**
Step I: Make groups of three digits from the right side. We have 064.
Step II: The first (and only) group is 64. Its unit digit is 4. The unit digit of a number's cube root is 4 only if the number itself ends in 4 (since \( 4^3 = 64 \)).
So, the cube root of 64 is 4.

(ii) 343
**Prime Factorization Method:**
First, we find the prime factors of 343:
\( 343 = 7 \times 7 \times 7 \)
Now, we group the factors in sets of three:
\( 343 = 7^3 \)
\( \implies \sqrt[3]{343} = 7 \)
**Estimation Method:**
Step I: Make groups of three digits from the right side. We have 343.
Step II: The first (and only) group is 343. Its unit digit is 3. The unit digit of a number's cube root is 7 only if the number itself ends in 3 (since \( 7^3 = 343 \)).
So, the cube root of 343 is 7.

(iii) 5832
**Prime Factorization Method:**
First, we break down 5832 into its prime factors:
\( 5832 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \)
Now, we group the factors in sets of three:
\( 5832 = (2 \times 2 \times 2) \times (3 \times 3 \times 3) \times (3 \times 3 \times 3) \)
\( 5832 = 2^3 \times 3^3 \times 3^3 \)
\( 5832 = (2 \times 3 \times 3)^3 \)
\( \implies \sqrt[3]{5832} = 2 \times 3 \times 3 = 18 \)
**Estimation Method:**
Step I: Make groups of three digits from the right side. We have 5 832.
Step II: The first group from the right is 832. Its unit digit is 2. The unit digit of the cube root will be 8 (since \( 8^3 = 512 \)).
Step III: The second group is 5. We need to find a number whose cube is close to 5 but not greater than 5. We know \( 1^3 = 1 \) and \( 2^3 = 8 \). Since \( 1^3 < 5 < 2^3 \), the tens digit of the cube root is 1.
Combining the digits, the cube root of 5832 is 18.

(iv) 74088
**Prime Factorization Method:**
First, we break down 74088 into its prime factors:
\( 74088 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 7 \times 7 \times 7 \)
Now, we group the factors in sets of three:
\( 74088 = (2 \times 2 \times 2) \times (3 \times 3 \times 3) \times (7 \times 7 \times 7) \)
\( 74088 = 2^3 \times 3^3 \times 7^3 \)
\( 74088 = (2 \times 3 \times 7)^3 \)
\( \implies \sqrt[3]{74088} = 2 \times 3 \times 7 = 42 \)
**Estimation Method:**
Step I: Make groups of three digits from the right side. We have 74 088.
Step II: The first group from the right is 088 (or simply 88). Its unit digit is 8. The unit digit of the cube root will be 2 (since \( 2^3 = 8 \)).
Step III: The second group is 74. We need to find a number whose cube is close to 74 but not greater than 74. We know \( 4^3 = 64 \) and \( 5^3 = 125 \). Since \( 4^3 < 74 < 5^3 \), the tens digit of the cube root is 4.
Combining the digits, the cube root of 74088 is 42.

(v) 3375
**Prime Factorization Method:**
First, we find the prime factors of 3375:
\( 3375 = 3 \times 3 \times 3 \times 5 \times 5 \times 5 \)
Now, we group the factors in sets of three:
\( 3375 = (3 \times 3 \times 3) \times (5 \times 5 \times 5) \)
\( 3375 = 3^3 \times 5^3 \)
\( 3375 = (3 \times 5)^3 \)
\( \implies \sqrt[3]{3375} = 3 \times 5 = 15 \)
**Estimation Method:**
Step I: Make groups of three digits from the right side. We have 3 375.
Step II: The first group from the right is 375. Its unit digit is 5. The unit digit of the cube root will be 5 (since \( 5^3 = 125 \)).
Step III: The second group is 3. We need to find a number whose cube is close to 3 but not greater than 3. We know \( 1^3 = 1 \) and \( 2^3 = 8 \). Since \( 1^3 < 3 < 2^3 \), the tens digit of the cube root is 1.
Combining the digits, the cube root of 3375 is 15.

(vi) 10648
**Prime Factorization Method:**
First, we break down 10648 into its prime factors:
\( 10648 = 2 \times 2 \times 2 \times 11 \times 11 \times 11 \)
Now, we group the factors in sets of three:
\( 10648 = (2 \times 2 \times 2) \times (11 \times 11 \times 11) \)
\( 10648 = 2^3 \times 11^3 \)
\( 10648 = (2 \times 11)^3 \)
\( \implies \sqrt[3]{10648} = 2 \times 11 = 22 \)
**Estimation Method:**
Step I: Make groups of three digits from the right side. We have 10 648.
Step II: The first group from the right is 648. Its unit digit is 8. The unit digit of the cube root will be 2 (since \( 2^3 = 8 \)).
Step III: The second group is 10. We need to find a number whose cube is close to 10 but not greater than 10. We know \( 2^3 = 8 \) and \( 3^3 = 27 \). Since \( 2^3 < 10 < 3^3 \), the tens digit of the cube root is 2.
Combining the digits, the cube root of 10648 is 22.

(vii) 46656
**Prime Factorization Method:**
First, we break down 46656 into its prime factors:
\( 46656 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \)
Now, we group the factors in sets of three:
\( 46656 = (2^3 \times 2^3) \times (3^3 \times 3^3) \)
\( 46656 = (2 \times 2 \times 3 \times 3)^3 \)
\( \implies \sqrt[3]{46656} = 2 \times 2 \times 3 \times 3 = 36 \)
**Estimation Method:**
Step I: Make groups of three digits from the right side. We have 46 656.
Step II: The first group from the right is 656. Its unit digit is 6. The unit digit of the cube root will be 6 (since \( 6^3 = 216 \)).
Step III: The second group is 46. We need to find a number whose cube is close to 46 but not greater than 46. We know \( 3^3 = 27 \) and \( 4^3 = 64 \). Since \( 3^3 < 46 < 4^3 \), the tens digit of the cube root is 3.
Combining the digits, the cube root of 46656 is 36.

(viii) 91125
**Prime Factorization Method:**
First, we break down 91125 into its prime factors:
\( 91125 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5 \)
Now, we group the factors in sets of three:
\( 91125 = (3 \times 3 \times 3) \times (3 \times 3 \times 3) \times (5 \times 5 \times 5) \)
\( 91125 = 3^3 \times 3^3 \times 5^3 \)
\( 91125 = (3 \times 3 \times 5)^3 \)
\( \implies \sqrt[3]{91125} = 3 \times 3 \times 5 = 45 \)
**Estimation Method:**
Step I: Make groups of three digits from the right side. We have 91 125.
Step II: The first group from the right is 125. Its unit digit is 5. The unit digit of the cube root will be 5 (since \( 5^3 = 125 \)).
Step III: The second group is 91. We need to find a number whose cube is close to 91 but not greater than 91. We know \( 4^3 = 64 \) and \( 5^3 = 125 \). Since \( 4^3 < 91 < 5^3 \), the tens digit of the cube root is 4.
Combining the digits, the cube root of 91125 is 45.
In simple words: To find the cube root, you can either break the number into prime factors and group them in threes, or use an estimation method by looking at the last digit and the remaining digits. Both methods help you find the number that, when multiplied by itself three times, gives the original number.

🎯 Exam Tip: When using the estimation method for cube roots, always remember the unit digits of cubes from 1 to 9, as they are unique. This helps quickly find the unit digit of the cube root.