GSEB Class 8 Maths Solutions Chapter 7 Cube and Cube Roots Exercise 7.2

Question 1. Find the cube root of each of the following numbers by the prime factorisation method.
(i) 64
(ii) 512
(iii) 10648
(iv) 27000
(v) 15625
(vi) 13824
(vii) 110592
(viii) 46656
(ix) 175616
(x) 91125
Answer:
(i) For 64, by prime factorization, we get:
\( 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \)
Grouping these into triplets:
\( 64 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \)
So, \( \sqrt[3]{64} = 2 \times 2 = 4 \)
Thus, the cube root of 64 is 4.

(ii) For 512, by prime factorization, we have:
\( 512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \)
Grouping these into triplets:
\( 512 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \)
So, \( \sqrt[3]{512} = 2 \times 2 \times 2 = 8 \)
Thus, the cube root of 512 is 8.

(iii) For 10648, by prime factorization, we have:
\( 10648 = 2 \times 2 \times 2 \times 11 \times 11 \times 11 \)
Grouping these into triplets:
\( 10648 = (2 \times 2 \times 2) \times (11 \times 11 \times 11) \)
So, \( \sqrt[3]{10648} = 2 \times 11 = 22 \)
Thus, the cube root of 10648 is 22.

(iv) For 27000, by prime factorization, we have:
\( 27000 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5 \)
Grouping these into triplets:
\( 27000 = (2 \times 2 \times 2) \times (3 \times 3 \times 3) \times (5 \times 5 \times 5) \)
So, \( \sqrt[3]{27000} = 2 \times 3 \times 5 = 30 \)
Thus, the cube root of 27000 is 30.

(v) For 15625, by prime factorization, we have:
\( 15625 = 5 \times 5 \times 5 \times 5 \times 5 \times 5 \)
Grouping these into triplets:
\( 15625 = (5 \times 5 \times 5) \times (5 \times 5 \times 5) \)
So, \( \sqrt[3]{15625} = 5 \times 5 = 25 \)
Thus, the cube root of 15625 is 25.

(vi) For 13824, by prime factorization, we have:
\( 13824 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \)
Grouping these into triplets:
\( 13824 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3) \)
So, \( \sqrt[3]{13824} = 2 \times 2 \times 2 \times 3 = 24 \)
Thus, the cube root of 13824 is 24.

(vii) For 110592, by prime factorization, we have:
\( 110592 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \)
Grouping these into triplets:
\( 110592 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3) \)
So, \( \sqrt[3]{110592} = 2 \times 2 \times 2 \times 2 \times 3 = 48 \)
Thus, the cube root of 110592 is 48.

(viii) For 46656, by prime factorization, we have:
\( 46656 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \)
Grouping these into triplets:
\( 46656 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3) \times (3 \times 3 \times 3) \)
So, \( \sqrt[3]{46656} = 2 \times 2 \times 3 \times 3 = 36 \)
Thus, the cube root of 46656 is 36.

(ix) For 175616, by prime factorization, we have:
\( 175616 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 7 \times 7 \times 7 \)
Grouping these into triplets:
\( 175616 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (7 \times 7 \times 7) \)
So, \( \sqrt[3]{175616} = 2 \times 2 \times 2 \times 7 = 56 \)
Thus, the cube root of 175616 is 56.

(x) For 91125, by prime factorization, we have:
\( 91125 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5 \)
Grouping these into triplets:
\( 91125 = (3 \times 3 \times 3) \times (3 \times 3 \times 3) \times (5 \times 5 \times 5) \)
So, \( \sqrt[3]{91125} = 3 \times 3 \times 5 = 45 \)
Thus, the cube root of 91125 is 45.
In simple words: To find the cube root of a number, first break it down into its smallest prime factors. Then, group these factors into sets of three identical numbers. For each group of three, you take one number. Multiply these single numbers together, and that product is your cube root.

Exam Tip: Always make sure to group prime factors into triplets when finding a cube root. If any factor is left over, the number is not a perfect cube.

Question 2. State true or false.
1. Cube of any odd number is even.
2. A perfect cube does not end with two zeros.
3. If the square of a number ends with 5, then its cube ends with 25.
4. There is no perfect cube which ends with 8.
5. The cube of a two-digit number may be a three-digit number.
6. The cube of a two-digit number may have seven or more digits.
7. The cube of a single-digit number may be a single-digit number.
Answer:
1. False
2. True
3. False
4. False
5. False
6. False
7. True
In simple words: We check each statement about cube properties. Odd numbers have odd cubes. Perfect cubes never have exactly two zeros at the end. A number's square ending in 5 does not mean its cube ends in 25 (e.g., \( 25^2 = 625 \), but \( 25^3 = 15625 \), which ends in 25, however \( 15^2=225 \) so \( 15^3=3375 \) which does not end in 25). Perfect cubes can end in 8 (e.g., \( 2^3 = 8 \)). The cube of a two-digit number will always have at least 4 digits, never just three. The cube of a two-digit number can have up to 6 digits, not seven or more. Finally, a single-digit number's cube can also be a single digit (e.g., \( 1^3 = 1 \), \( 2^3 = 8 \)).

Exam Tip: To answer true/false questions about number properties, it is often helpful to test a few small examples for each statement.

Question 3. You know that 1,331 is a perfect cube. Can you guess without factorisation what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768?
Answer:
1. For 1331:
We separate the number 1331 into two groups: 1 and 331.
The unit digit of 331 is 1. Since \( 1^3 = 1 \), the unit digit of the cube root must be 1.
Now, consider the first group, 1. The largest perfect cube less than or equal to 1 is \( 1^3 = 1 \). So, the tens digit of the cube root is 1.
Thus, the cube root of 1331 is 11.

2. For 4913:
We separate the number 4913 into two groups: 4 and 913.
The unit digit of 913 is 3. Since numbers ending in 3 have cubes ending in 7 (because \( 7^3 = 343 \)), the unit digit of the cube root must be 7.
Now, consider the first group, 4. The largest perfect cube less than or equal to 4 is \( 1^3 = 1 \). (\( 2^3 = 8 \), which is greater than 4). So, the tens digit of the cube root is 1.
Thus, the cube root of 4913 is 17.

3. For 12167:
We separate the number 12167 into two groups: 12 and 167.
The unit digit of 167 is 7. Since numbers ending in 7 have cubes ending in 3 (because \( 3^3 = 27 \)), the unit digit of the cube root must be 3.
Now, consider the first group, 12. The largest perfect cube less than or equal to 12 is \( 2^3 = 8 \). (\( 3^3 = 27 \), which is greater than 12). So, the tens digit of the cube root is 2.
Thus, the cube root of 12167 is 23.

4. For 32768:
We separate the number 32768 into two groups: 32 and 768.
The unit digit of 768 is 8. Since numbers ending in 8 have cubes ending in 2 (because \( 2^3 = 8 \)), the unit digit of the cube root must be 2.
Now, consider the first group, 32. The largest perfect cube less than or equal to 32 is \( 3^3 = 27 \). (\( 4^3 = 64 \), which is greater than 32). So, the tens digit of the cube root is 3.
Thus, the cube root of 32768 is 32.
In simple words: To guess a cube root without factoring, you split the number into two parts. The last group tells you the last digit of the root (e.g., if it ends in 1, the root ends in 1; if it ends in 3, the root ends in 7). The first group tells you the first digit of the root by finding the biggest perfect cube that fits inside it.

Exam Tip: Memorize the unit digits of cubes from 1 to 10 to quickly determine the unit digit of a cube root without factorization.