Welcome dear students! Today we are going to learn about Cubes and Cube Roots from Class 8 Maths.
This is a story about one of India's great mathematical geniuses, S. Ramanujan. Once another famous mathematician Professor G.H. Hardy came to visit him in a taxi whose number was 1729. While talking to Ramanujan, Hardy described this number as a dull number. Ramanujan quickly pointed out that 1729 was indeed interesting. He said it is the smallest number that can be expressed as a sum of two cubes in two different ways: 1729 = 1728 + 1 = 12³ + 1³. 1729 = 1000 + 729 = 10³ + 9³. 1729 has since been known as the Hardy–Ramanujan Number, even though this feature of 1729 was known more than 300 years before Ramanujan. How did Ramanujan know this? Well, he loved numbers. All through his life, he experimented with numbers. He probably found numbers that were expressed as the sum of two squares and sum of two cubes also. There are many other interesting patterns of cubes. Let us learn about cubes, cube roots and many other interesting facts related to them.
The Hardy–Ramanujan Number 1729 is the smallest Hardy–Ramanujan Number. There are infinitely many such numbers. A few are 4104, which can be written as 2³ + 16³ and 9³ + 15³. Another is 13832, which can be written as 18³ + 20³ and 2³ + 24³. You can check these with the numbers given in the brackets.
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You know that the word cube is used in geometry. A cube is a solid figure which has all its sides equal. How many cubes of side 1 cm will make a cube of side 2 cm? How many cubes of side 1 cm will make a cube of side 3 cm? Consider the numbers 1, 8, 27, and so on. These are called perfect cubes or cube numbers. Can you say why they are named so? Each of them is obtained when a number is multiplied by taking it three times. Figures which have 3-dimensions are known as solid figures. We note that 1 = 1 × 1 × 1 = 1³; 8 = 2 × 2 × 2 = 2³; 27 = 3 × 3 × 3 = 3³. Since 5³ = 5 × 5 × 5 = 125, therefore 125 is a cube number. Is 9 a cube number? No, as 9 = 3 × 3 and there is no natural number which multiplied by taking three times gives 9. We can see also that 2 × 2 × 2 = 8 and 3 × 3 × 3 = 27. This shows that 9 is not a perfect cube.
Let us look at the cubes of numbers from 1 to 10. 1³ = 1. 2³ = 8. 3³ = 27. 4³ = 64. 5³ = 125. 6³ = 216. 7³ = 343. 8³ = 512. 9³ = 729. 10³ = 1000. The numbers 729, 1000, and 1728 are also perfect cubes. There are only ten perfect cubes from 1 to 1000. You can check this. How many perfect cubes are there from 1 to 100? Observe the cubes of even numbers. Are they all even? What can you say about the cubes of odd numbers? Following are the cubes of the numbers from 11 to 20. 11³ = 1331. 12³ = 1728. 13³ = 2197. 14³ = 2744. 15³ = 3375. 16³ = 4096. 17³ = 4913. 18³ = 5832. 19³ = 6859. 20³ = 8000. Even numbers have even cubes, and odd numbers have odd cubes.
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Consider a few numbers having 1 as the ones digit. Find the cube of each of them. What can you say about the ones digit of the cube of a number having 1 as the ones digit? Similarly, explore the ones digit of cubes of numbers ending in 2, 3, 4, and so on.
TRY THESE: Find the ones digit of the cube of each of the following numbers. For 3331, the ones digit is 1, so its cube ends in 1. For 8888, the ones digit is 8, and 8³ = 512, so it ends in 2. For 149, the ones digit is 9, and 9³ = 729, so it ends in 9. For 1005, the ones digit is 5, and 5³ = 125, so it ends in 5. For 1024, the ones digit is 4, and 4³ = 64, so it ends in 4. For 77, the ones digit is 7, and 7³ = 343, so it ends in 3. For 5022, the ones digit is 2, and 2³ = 8, so it ends in 8. For 53, the ones digit is 3, and 3³ = 27, so it ends in 7.
Now let us move on to some interesting patterns. First, adding consecutive odd numbers. Observe the following pattern of sums of odd numbers. 1 = 1 = 1³. 3 + 5 = 8 = 2³. 7 + 9 + 11 = 27 = 3³. 13 + 15 + 17 + 19 = 64 = 4³. 21 + 23 + 25 + 27 + 29 = 125 = 5³. Is it not interesting? How many consecutive odd numbers will be needed to obtain the sum as 10³? You will need ten consecutive odd numbers. The sequence starts from 91. So, 91 + 93 + 95 + 97 + 99 + 101 + 103 + 105 + 107 + 109 = 1000.
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TRY THESE: Express the following numbers as the sum of odd numbers using the above pattern. For 6³, we need six consecutive odd numbers. The sequence starts at 31. So, 31 + 33 + 35 + 37 + 39 + 41 = 216. For 8³, we need eight consecutive odd numbers. They are 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71 = 512. For 7³, we need seven consecutive odd numbers. They are 43 + 45 + 47 + 49 + 51 + 53 + 55 = 343.
Consider the following pattern. 2³ - 1³ = 1 + 2 × 1 × 3. 3³ - 2³ = 1 + 3 × 2 × 3. 4³ - 3³ = 1 + 4 × 3 × 3. Using this pattern, find the value of the following. 7³ - 6³ = 1 + 7 × 6 × 3 = 1 + 126 = 127. 12³ - 11³ = 1 + 12 × 11 × 3 = 1 + 396 = 397. 20³ - 19³ = 1 + 20 × 19 × 3 = 1 + 1140 = 1141. 51³ - 50³ = 1 + 51 × 50 × 3 = 1 + 7650 = 7651.
Second pattern: Cubes and their prime factors. Consider the prime factorisation of numbers and their cubes. 4 = 2 × 2, and 4³ = 64 = 2 × 2 × 2 × 2 × 2 × 2 = 2³ × 2³. 6 = 2 × 3, and 6³ = 216 = 2 × 2 × 2 × 3 × 3 × 3 = 2³ × 3³. 15 = 3 × 5, and 15³ = 3375 = 3 × 3 × 3 × 5 × 5 × 5 = 3³ × 5³. 12 = 2 × 2 × 3, and 12³ = 1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 = 2³ × 2³ × 3³. Observe that each prime factor of a number appears three times in the prime factorisation of its cube.
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In the prime factorisation of any number, if each factor appears three times, then the number is a perfect cube. Think about it. Is 216 a perfect cube? By prime factorisation, 216 = 2 × 2 × 2 × 3 × 3 × 3. Each factor appears 3 times. 216 = 2³ × 3³ = (2 × 3)³ = 6³, which is a perfect cube. Is 729 a perfect cube? 729 = 3 × 3 × 3 × 3 × 3 × 3. Yes, 729 is a perfect cube. Now let us check for 500. Prime factorisation of 500 is 2 × 2 × 5 × 5 × 5. So, 500 is not a perfect cube.
Let us solve Example 1. Is 243 a perfect cube? Solution: 243 = 3 × 3 × 3 × 3 × 3. In the above factorisation 3 × 3 remains after grouping the 3's in triplets. Therefore, 243 is not a perfect cube.
TRY THESE: Which of the following are perfect cubes? 1. 400. Prime factorisation is 2 × 2 × 2 × 2 × 5 × 5. Not all factors appear three times. Not a perfect cube. 2. 3375. Prime factorisation is 3 × 3 × 3 × 5 × 5 × 5. All appear in triplets. It is a perfect cube. 3. 8000. Prime factorisation is 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5. It is a perfect cube. 4. 15625. Prime factorisation is 5 × 5 × 5 × 5 × 5 × 5. It is a perfect cube. 5. 9000. Prime factorisation is 2 × 2 × 2 × 3 × 3 × 5 × 5 × 5. Not a perfect cube. 6. 6859. Prime factorisation is 19 × 19 × 19. It is a perfect cube. 7. 2025. Prime factorisation is 3 × 3 × 3 × 3 × 5 × 5. Not a perfect cube. 8. 10648. Prime factorisation is 2 × 2 × 2 × 11 × 11 × 11. It is a perfect cube.
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Now, let us learn about the smallest multiple that is a perfect cube. Raj made a cuboid of plasticine. Length, breadth and height of the cuboid are 15 cm, 30 cm, 15 cm respectively. Anu asks how many such cuboids will she need to make a perfect cube. Can you tell? Raj said, Volume of cuboid is 15 × 30 × 15 = 3 × 5 × 2 × 3 × 5 × 3 × 5 = 2 × 3 × 3 × 3 × 5 × 5 × 5. Since there is only one 2 in the prime factorisation, we need 2 × 2, that is 4, to make it a perfect cube. Therefore, we need 4 such cuboids to make a cube.
Example 2: Is 392 a perfect cube? If not, find the smallest natural number by which 392 must be multiplied so that the product is a perfect cube. Solution: 392 = 2 × 2 × 2 × 7 × 7. The prime factor 7 does not appear in a group of three. Therefore, 392 is not a perfect cube. To make it a cube, we need one more 7. In that case, 392 × 7 = 2 × 2 × 2 × 7 × 7 × 7 = 2744, which is a perfect cube. Hence the smallest natural number by which 392 should be multiplied to make a perfect cube is 7.
Example 3: Is 53240 a perfect cube? If not, then by which smallest natural number should 53240 be divided so that the quotient is a perfect cube? Solution: 53240 = 2 × 2 × 2 × 11 × 11 × 11 × 5. The prime factor 5 does not appear in a group of three. So, 53240 is not a perfect cube. In the factorisation, 5 appears only one time. If we divide the number by 5, then the prime factorisation of the quotient will not contain 5. So, 53240 ÷ 5 = 2 × 2 × 2 × 11 × 11 × 11. Hence the smallest number by which 53240 should be divided to make it a perfect cube is 5. The perfect cube in that case is 10648.
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Example 4: Is 1188 a perfect cube? If not, by which smallest natural number should 1188 be divided so that the quotient is a perfect cube? Solution: 1188 = 2 × 2 × 3 × 3 × 3 × 11. The primes 2 and 11 do not appear in groups of three. So, 1188 is not a perfect cube. In the factorisation of 1188 the prime 2 appears only two times and the prime 11 appears once. So, if we divide 1188 by 2 × 2 × 11 = 44, then the prime factorisation of the quotient will not contain 2 and 11. Hence the smallest natural number by which 1188 should be divided to make it a perfect cube is 44. And the resulting perfect cube is 1188 ÷ 44 = 27 = 3³.
Example 5: Is 68600 a perfect cube? If not, find the smallest number by which 68600 must be multiplied to get a perfect cube. Solution: We have, 68600 = 2 × 2 × 2 × 5 × 5 × 7 × 7 × 7. In this factorisation, we find that there is no triplet of 5. So, 68600 is not a perfect cube. To make it a perfect cube we multiply it by 5. Thus, 68600 × 5 = 2 × 2 × 2 × 5 × 5 × 5 × 7 × 7 × 7 = 343000, which is a perfect cube. Observe that 343 is a perfect cube. From Example 5 we know that 343000 is also a perfect cube.
THINK, DISCUSS AND WRITE: Check which of the following are perfect cubes. 2700 is not a perfect cube. 16000 is not a perfect cube. 64000 is a perfect cube. 900 is not a perfect cube. 125000 is a perfect cube. 36000 is not a perfect cube. 21600 is not a perfect cube. 10000 is not a perfect cube. 27000000 is a perfect cube. 1000 is a perfect cube. What pattern do you observe in these perfect cubes? You will observe that perfect cubes ending in zeros always have a number of zeros that is a multiple of three.
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Let us solve Exercise 6.1. Question 1: Which of the following numbers are not perfect cubes? i. 216. Prime factors are 2³ × 3³. It is a perfect cube. ii. 128. Prime factors are 2⁷. Not a multiple of three. Not a perfect cube. iii. 1000. Prime factors are 2³ × 5³. It is a perfect cube. iv. 100. Prime factors are 2² × 5². Not a perfect cube. v. 46656. Prime factors are 2⁶ × 3⁶. It is a perfect cube. So, 128 and 100 are not perfect cubes.
Question 2: Find the smallest number by which each must be multiplied to obtain a perfect cube. i. 243. Factors: 3⁵. Needs one more 3. Multiply by 3. ii. 256. Factors: 2⁸. Needs one more 2. Multiply by 2. iii. 72. Factors: 2³ × 3². Needs one more 3. Multiply by 3. iv. 675. Factors: 3³ × 5². Needs one more 5. Multiply by 5. v. 100. Factors: 2² × 5². Needs one more 2 and one more 5. Multiply by 10.
Question 3: Find the smallest number by which each must be divided to obtain a perfect cube. i. 81. Factors: 3⁴. Divide by 3. ii. 128. Factors: 2⁷. Divide by 2. iii. 135. Factors: 3³ × 5. Divide by 5. iv. 192. Factors: 2⁶ × 3. Divide by 3. v. 704. Factors: 2⁶ × 11. Divide by 11.
Question 4: Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube? Volume is 5 × 2 × 5 = 50. Prime factors are 2 × 5². To make a cube, we need 2² × 5, which is 20. So, he needs 20 such cuboids.
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Now let us move to section 6.3, Cube Roots. If the volume of a cube is 125 cm³, what would be the length of its side? To get the length of the side of the cube, we need to know a number whose cube is 125. Finding the square root, as you know, is the inverse operation of squaring. Similarly, finding the cube root is the inverse operation of finding cube. We know that 2³ = 8, so we say that the cube root of 8 is 2. We write cbrt(8) = 2. The symbol cbrt denotes cube root. Consider the following relationships. 1³ = 1, so cbrt(1) = 1. 6³ = 216, so cbrt(216) = 6. 2³ = 8, so cbrt(8) = cbrt(2³) = 2. 7³ = 343, so cbrt(343) = 7. 3³ = 27, so cbrt(27) = cbrt(3³) = 3. 8³ = 512, so cbrt(512) = 8. 4³ = 64, so cbrt(64) = 4. 9³ = 729, so cbrt(729) = 9. 5³ = 125, so cbrt(125) = 5. 10³ = 1000, so cbrt(1000) = 10.
Now, 6.3.1: Cube root through prime factorisation method. Consider 3375. We find its cube root by prime factorisation: 3375 = 3 × 3 × 3 × 5 × 5 × 5 = 3³ × 5³ = (3 × 5)³. Therefore, cube root of 3375 = cbrt(3375) = 3 × 5 = 15. Similarly, to find cbrt(74088), we have, 74088 = 2 × 2 × 2 × 3 × 3 × 3 × 7 × 7 × 7 = 2³ × 3³ × 7³ = (2 × 3 × 7)³. Therefore, cbrt(74088) = 2 × 3 × 7 = 42.
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Example 6: Find the cube root of 8000. Solution: Prime factorisation of 8000 is 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5. So, cbrt(8000) = 2 × 2 × 5 = 20.
Example 7: Find the cube root of 13824 by prime factorisation method. Solution: 13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 = 2³ × 2³ × 2³ × 3³. Therefore, cbrt(13824) = 2 × 2 × 2 × 3 = 24.
THINK, DISCUSS AND WRITE: State true or false: for any integer m, m² < m³. Why? This is false. For example, if m = 1, 1² = 1 and 1³ = 1, so they are equal. If m = 0, both are zero. If m is negative, say -1, (-1)² = 1, and (-1)³ = -1, so m² > m³. So the statement is not true for all integers.
Let us solve Exercise 6.2. Question 1: Find the cube root by prime factorisation method. i. 64. Factors: 2³ × 2³. Cube root is 4. ii. 512. Factors: 2³ × 2³ × 2³. Cube root is 8. iii. 10648. Factors: 2³ × 11³. Cube root is 22. iv. 27000. Factors: 2³ × 3³ × 5³. Cube root is 30. v. 15625. Factors: 5³ × 5³. Cube root is 25. vi. 13824. Factors: 2³ × 2³ × 2³ × 3³. Cube root is 24. vii. 110592. Factors: 2³ × 2³ × 2³ × 2³ × 3³. Cube root is 48. viii. 46656. Factors: 2³ × 2³ × 3³ × 3³. Cube root is 36. ix. 175616. Factors: 2³ × 2³ × 2³ × 7³. Cube root is 56. x. 91125. Factors: 3³ × 3³ × 5³. Cube root is 45.
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Question 2: State true or false. i. Cube of any odd number is even. False. Odd × odd × odd is odd. ii. A perfect cube does not end with two zeros. True. It ends with a multiple of three zeros. iii. If square of a number ends with 5, then its cube ends with 25. False. For example, 15² = 225, but 15³ = 3375, which ends in 75. iv. There is no perfect cube which ends with 8. False. 2³ = 8, and 12³ = 1728. v. The cube of a two digit number may be a three digit number. False. Smallest two digit is 10, 10³ = 1000, which is four digits. vi. The cube of a two digit number may have seven or more digits. False. Largest two digit is 99, 99³ = 970299, which is six digits. vii. The cube of a single digit number may be a single digit number. True. 1³ = 1, and 2³ = 8.
Finally, let us review what we have discussed. One: Numbers like 1729, 4104, 13832, are known as Hardy–Ramanujan Numbers. They can be expressed as sum of two cubes in two different ways. Two: Numbers obtained when a number is multiplied by itself three times are known as cube numbers. For example 1, 8, 27, and so on. Three: If in the prime factorisation of any number each factor appears three times, then the number is a perfect cube. Four: The symbol cbrt denotes cube root. For example cbrt(27) = 3. Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]