RBSE Solutions Class 8 Maths Chapter 2 Cube and Cube Roots Exercise 2.1

Question 1. Which of the following numbers are not perfect cube?
(i) 512
(ii) 243
(iii) 1000
(iv) 100
(v) 2700
Answer:
(i) For 512:
The prime factors of 512 are \( 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \), which can be written as \( 2^9 \). This can be grouped as \( (2^3) \times (2^3) \times (2^3) \). Since all prime factors form complete groups of three, 512 is a perfect cube. A perfect cube is a number that can be made by multiplying an integer by itself three times.
(ii) For 243:
The prime factors of 243 are \( 3 \times 3 \times 3 \times 3 \times 3 \), which can be written as \( 3^3 \times 3^2 \). Here, the factor 3 appears five times. While three of them form a group (\( 3^3 \)), two 3s are left over (\( 3^2 \)). Since all prime factors do not form complete groups of three, 243 is not a perfect cube.
(iii) For 1000:
The prime factors of 1000 are \( 2 \times 2 \times 2 \times 5 \times 5 \times 5 \), which can be written as \( 2^3 \times 5^3 \). Both the factors 2 and 5 form complete groups of three. Therefore, 1000 is a perfect cube. This means \( 1000 = (2 \times 5)^3 = 10^3 \).
(iv) For 100:
The prime factors of 100 are \( 2 \times 2 \times 5 \times 5 \), which can be written as \( 2^2 \times 5^2 \). Neither the factor 2 nor the factor 5 appears in groups of three. So, 100 is not a perfect cube. Numbers like 100 are perfect squares (\( 10^2 \)) but not perfect cubes.
(v) For 2700:
The prime factors of 2700 are \( 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 5 \), which can be written as \( 2^2 \times 3^3 \times 5^2 \). While the factor 3 forms a perfect group of three, the factors 2 and 5 do not. Therefore, 2700 is not a perfect cube. To become a perfect cube, 2700 would need one more 2 and one more 5 as factors.
In simple words: A perfect cube is a number where all its prime factors can be grouped into sets of three. 512 and 1000 are perfect cubes because their factors form complete groups of three. However, 243, 100, and 2700 are not perfect cubes because some of their prime factors do not form complete groups of three.

🎯 Exam Tip: To check if a number is a perfect cube, always perform prime factorization and see if all prime factors can be arranged into sets of three. If even one factor does not form a complete triplet, the number is not a perfect cube.

Question 2. Find out the smallest number multiplied by the following numbers to get the perfect cube?
(i) 108
(ii) 500
(iii) 5400
(iv) 10584
Answer:
(i) For 108:
First, we find the prime factors of 108, which are \( 2 \times 2 \times 3 \times 3 \times 3 \). We can see that the factor 3 forms a perfect group of three (\( 3^3 \)). However, the factor 2 only appears twice (\( 2^2 \)). To make it a perfect cube, we need one more 2 to complete the group of three. So, the smallest number to multiply by is 2. Multiplying 108 by 2 makes it 216, which is \( 6^3 \).
(ii) For 500:
To find the smallest number to multiply by, we first find the prime factors of 500. These are \( 2 \times 2 \times 5 \times 5 \times 5 \). The factor 5 is already in a group of three (\( 5^3 \)), but the factor 2 appears only twice (\( 2^2 \)). To complete the group of three for factor 2, we need one more 2. Therefore, we should multiply by 2. This will change the number to \( 2^3 \times 5^3 \), which is \( (2 \times 5)^3 = 10^3 = 1000 \).
(iii) For 5400:
Let's find the prime factors of 5400. They are \( 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 5 \). We see that 2s and 3s form complete groups of three (\( 2^3 \) and \( 3^3 \)). However, the factor 5 only appears twice (\( 5^2 \)). To complete its group of three, we need one more 5. So, the smallest number to multiply by is 5. This multiplication would result in \( (2 \times 3 \times 5)^3 = 30^3 = 27000 \).
(iv) For 10584:
To determine the smallest multiplier, we first resolve 10584 into its prime factors. These are \( 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 7 \times 7 \). Here, the factors 2 and 3 already form complete groups of three (\( 2^3 \) and \( 3^3 \)). However, the factor 7 appears only twice (\( 7^2 \)). To complete its group, we need one more 7. Therefore, the smallest number by which to multiply is 7. After multiplying by 7, the number becomes \( (2 \times 3 \times 7)^3 = 42^3 = 74088 \).
In simple words: To make a number a perfect cube by multiplying, first find its prime factors. For any factor that doesn't have a group of three, count how many more are needed to complete the group. Multiply the original number by that missing factor(s).

🎯 Exam Tip: When finding the smallest number to multiply, only focus on the prime factors that are not in complete triplets. For each such factor, multiply by itself enough times to complete a set of three.

Question 3. Find out the smallest number by which following numbers must be divided to get the perfect cube.
(i) 24
(ii) 250
(iii) 192
(iv) 135
Answer:
(i) For 24:
First, we find the prime factors of 24, which are \( 2 \times 2 \times 2 \times 3 \). We see that the factor 2 forms a perfect group of three (\( 2^3 \)). However, the factor 3 appears only once. To make 24 a perfect cube, we need to remove the extra factor 3. So, the smallest number to divide by is 3. Dividing 24 by 3 gives 8, which is \( 2^3 \).
(ii) For 250:
To find the smallest number to divide by, we first find the prime factors of 250. These are \( 2 \times 5 \times 5 \times 5 \). The factor 5 is already in a perfect group of three (\( 5^3 \)). However, the factor 2 appears only once. To make 250 a perfect cube, we need to remove this extra factor 2. Therefore, we should divide by 2. Dividing 250 by 2 gives 125, which is \( 5^3 \).
(iii) For 192:
Let's find the prime factors of 192. They are \( 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \). We see that the factor 2 forms two complete groups of three (\( 2^3 \times 2^3 = 2^6 \)). However, the factor 3 appears only once. To make 192 a perfect cube, we need to remove this extra factor 3. So, the smallest number to divide by is 3. After dividing by 3, the number becomes 64, which is \( 4^3 \).
(iv) For 135:
To determine the smallest divisor, we first resolve 135 into its prime factors. These are \( 3 \times 3 \times 3 \times 5 \). Here, the factor 3 already forms a complete group of three (\( 3^3 \)). However, the factor 5 appears only once. To make 135 a perfect cube, we need to remove this extra factor 5. Therefore, the smallest number by which to divide is 5. Dividing 135 by 5 results in 27, which is \( 3^3 \).
In simple words: To make a number a perfect cube by dividing, find its prime factors. Identify any factor that does not have a group of three. Divide the original number by that extra factor(s) to remove it.

🎯 Exam Tip: When finding the smallest number to divide, identify all prime factors that do not form complete triplets. The product of these 'extra' factors is the smallest number by which the original number should be divided.