GSEB Class 8 Maths Solutions Chapter 7 Cube and Cube Roots Exercise 7.1

Question 1. Which of the following numbers are not perfect cubes?
1. 216
2. 128
3. 1000
4. 100
5. 46656
Answer:
1. We find the prime factors of 216. \( 216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \). When we group these prime factors into sets of three, no factor remains ungrouped. Therefore, 216 is a perfect cube.
2. We find the prime factors of 128. \( 128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \). After grouping the prime factors of 128 into sets of three, we are left with one factor of 2 that is not part of a triple. Consequently, 128 is not a perfect cube.
3. We determine the prime factors of 1000. \( 1000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5 \). When we group these prime factors into sets of three, no factor is left over. Hence, 1000 is a perfect cube.
4. We find the prime factors of 100. \( 100 = 2 \times 2 \times 5 \times 5 \). When grouping the prime factors into sets of three, we do not obtain any complete triples. The factors \( 2 \times 2 \) and \( 5 \times 5 \) are not in triples. Thus, 100 is not a perfect cube.
5. We determine the prime factors of 46656. \( 46656 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \). Grouping these prime factors into sets of three, we find that no prime factor is left over. Therefore, 46656 is a perfect cube.
The numbers that are not perfect cubes are 128 and 100.
In simple words: To check if a number is a perfect cube, break it down into its prime factors. If you can group all factors into sets of three, it's a perfect cube. If any factor is left over, it's not. From the given numbers, 128 and 100 are not perfect cubes because their prime factors cannot all be grouped into sets of three.

Exam Tip: Always perform prime factorization and look for complete triples of factors to quickly identify perfect cubes or numbers that are not perfect cubes.

Question 2. Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube?
1. 243
2. 256
3. 72
4. 675
5. 100
Answer:
1. We start by finding the prime factors of 243. \( 243 = 3 \times 3 \times 3 \times 3 \times 3 \). Here, the prime factor 3 appears five times; one '3' is not part of a group of three. This means 243 is not a perfect cube. To make it a perfect cube, we need one more factor of 3. So, \( 243 \times 3 = 729 \), and \( 729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \). Thus, the smallest number needed to multiply 243 by to achieve a perfect cube is 3.
2. We begin by finding the prime factors of 256. \( 256 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \). When grouping the prime factors of 256 into sets of three, we are left with two factors of 2 (i.e., \( 2 \times 2 \)) that are not part of a triple. This indicates 256 is not a perfect cube. To make it a perfect cube, we require one more factor of 2. So, \( 256 \times 2 = 512 \), and \( 512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \). Therefore, the smallest number needed to multiply 256 by to obtain a perfect cube is 2.
3. We find the prime factors of 72. \( 72 = 2 \times 2 \times 2 \times 3 \times 3 \). Grouping the prime factors of 72 into sets of three, we find that the factors \( 3 \times 3 \) are left over, not forming a complete triple. This shows 72 is not a perfect cube. To achieve a perfect cube, we need one more factor of 3. So, \( 72 \times 3 = 216 \), and \( 216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \). Hence, the smallest number by which 72 must be multiplied to make it a perfect cube is 3.
4. We determine the prime factors of 675. \( 675 = 3 \times 3 \times 3 \times 5 \times 5 \). When grouping the prime factors of 675 into sets of three, we are left with factors \( 5 \times 5 \), which do not form a complete triple. Therefore, 675 is not a perfect cube. To create a perfect cube, we need one more factor of 5. So, \( 675 \times 5 = 3375 \), and \( 3375 = 3 \times 3 \times 3 \times 5 \times 5 \times 5 \). As a result, the smallest number by which 675 must be multiplied to make it a perfect cube is 5.
5. We find the prime factors of 100. \( 100 = 2 \times 2 \times 5 \times 5 \). The prime factors are not in groups of three. This means 100 is not a perfect cube. To make it a perfect cube, we need one more factor of 2 and one more factor of 5. So, we multiply by \( 2 \times 5 = 10 \). Hence, \( 100 \times 10 = 1000 \), and \( 1000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5 \). Thus, 1000 becomes a perfect cube, and the smallest number required is 10.
In simple words: To find the smallest number to multiply by, first get the prime factors of the given number. Look for any factors that are not in groups of three. Multiply the original number by whatever factors are needed to complete all the triples. This will give you the smallest number required.

Exam Tip: Prime factorization is key! Always list out all prime factors and clearly show which factors are incomplete triples before identifying the multiplier.

Question 3. Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.
1. 81
2. 128
3. 135
4. 192
5. 704
Answer:
1. We begin by finding the prime factors of 81. \( 81 = 3 \times 3 \times 3 \times 3 \). When grouping the prime factors into sets of three, we are left with one factor of 3. This means 81 is not a perfect cube. To achieve a perfect cube, we need to divide 81 by this leftover factor of 3. So, \( 81 \div 3 = 27 \), and \( 27 = 3 \times 3 \times 3 \). Thus, 27 is a perfect cube, and the smallest number required to divide 81 by is 3.
2. We find the prime factors of 128. \( 128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \). Grouping the prime factors of 128 into sets of three, we are left with one factor of 2. This shows 128 is not a perfect cube. To make it a perfect cube, we must divide 128 by this leftover factor of 2. So, \( 128 \div 2 = 64 \), and \( 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \). Hence, 64 is a perfect cube, and the smallest number required is 2.
3. We determine the prime factors of 135. \( 135 = 3 \times 3 \times 3 \times 5 \). When grouping the prime factors of 135 into sets of three, we are left with one factor of 5. This indicates 135 is not a perfect cube. To achieve a perfect cube, we should divide 135 by this leftover factor of 5. So, \( 135 \div 5 = 27 \), and \( 27 = 3 \times 3 \times 3 \). Therefore, 27 is a perfect cube, and the smallest number required is 5.
4. We find the prime factors of 192. \( 192 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \). Grouping the prime factors of 192 into sets of three, we find that one factor of 3 is left over. This shows 192 is not a perfect cube. To make it a perfect cube, we must divide 192 by this leftover factor of 3. So, \( 192 \div 3 = 64 \), and \( 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \). Hence, 64 is a perfect cube, and the smallest number required is 3.
5. We determine the prime factors of 704. \( 704 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 11 \). When grouping the prime factors of 704 into sets of three, we are left with one factor of 11. This indicates 704 is not a perfect cube. To form a perfect cube, we should divide 704 by this leftover factor of 11. So, \( 704 \div 11 = 64 \), and \( 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \). Thus, 64 is a perfect cube, and the smallest number required is 11.
In simple words: To find the smallest number to divide by, first find all the prime factors of the given number. Identify any factors that do not form a complete group of three. The product of these 'leftover' factors is the smallest number you need to divide by to get a perfect cube.

Exam Tip: Be careful to identify *all* leftover factors for division. If you have \( 2 \times 2 \) leftover, you divide by 4. If you have \( 3 \times 3 \) leftover, you divide by 9.

Question 4. Pariksht makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?
Answer: The sides of the plasticine cuboid are 5 cm, 2 cm, and 5 cm. To create a perfect cube from these cuboids, each dimension (length, width, height) must have prime factors that form triples. The existing dimensions are \( 5 \times 5 \times 2 \). To make each factor a triple, we need one more 5, two more 2s (i.e., \( 2 \times 2 \)). So, we need to multiply the dimensions by \( 5 \times 2 \times 2 \). The factors for the cube should be \( (5 \times 5 \times 5) \times (2 \times 2 \times 2) \). The current factors are \( 5 \times 5 \times 2 \). We need to multiply by \( 5 \times 2 \times 2 = 20 \). Therefore, Pariksht will need 20 such cuboids to form a complete cube.
In simple words: The cuboid has sides 5, 2, and 5. To make a big cube, all sides must have factors in groups of three. For the number 5, we have two ( \( 5 \times 5 \) ), so we need one more 5. For the number 2, we have one ( \( 2 \) ), so we need two more 2s ( \( 2 \times 2 \) ). Multiplying these missing factors \( (5 \times 2 \times 2) \) gives us 20. So, 20 small cuboids are needed to build a big cube.

Exam Tip: For forming a cube from cuboids, ensure all prime factors of the dimensions are in groups of three. The number of additional cuboids needed is the product of the missing factors required to complete these triples.