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Cube
Cube of a number:When a number is multiplied three times by itself, the product obtained is called the cube of a number.For a given numberm, we define, cube of m = m × m × m, denoted by m^{3}. For example: (i) 2^{3} = (2 × 2 × 2) = 8. Thus, cube of 2 is 8. (ii) 3^{3} = (3 × 3 × 3) = 27. Thus, cube of 3 is 27. (iii)4 × 4 × 4 = 64, here 64 is the cube of 4. (iv) 5 × 5 × 5 = 125, here 125 is the cube of 5. Perfect cube:A natural number (n) is said to be a perfect cube if (n = m^{3}) it is the cube of some natural number.For example: 1^{3} =1, 2^{3} = 8, 3^{3} =27, 4^{3} =64, 5^{3} =125, etc. Thus 1, 8, 27, 64, 125, etc. are perfect cubes. A given natural number is a perfect cube if it can be expressed as the product of triplets of equal factors. Cubes of negative integer:The cube of a negative integer is always negative.For example: (1)^{3} = (1) × (1) × (1) = 1, (2)^{3} = (2) × (2) × (2) = 8 (3)^{3} = (3) × (3) × (3) = 27, etc. Cube of a rational number:We have, (a/b) ^{3} = a/b × a/b × a/b = (a × a × a)/(b × b × b) = a^{3}/b^{3}Hence, (a/b) ^{3} = a^{3}/ b^{3} For example: (i) (3/5) ^{3} = 3^{3}/5 ^{3} = (3 × 3 × 3)/(5 × 5 × 5) = 27/125 (ii) (2/3) ^{3} = (2) ^{3}/ 3^{3} = {(2) × (2) × (2)}/(3 × 3 × 3) = 8/27 Properties of cubes of numbers:(i) The cube of every even natural number is even.(ii) The cube of every odd natural number is odd. Solved example to find perfect cubes step by step;1. Show that 189 is not a perfect cube.Solution: Resolving 189 into prime factors, we get: 189 = 3 × 3 × 3 × 7 Making triplets, we find that one triplet is formed and we are left with one more factor. Thus, 189 cannot be expressed as a product of triplets. Hence, 189 is not a perfect cube. 2. Show that 216 is a perfect cube. Find the number whose cube is 216. Solution: Resolving 216 into prime factors, we get: 216 = 2 × 2 × 2 × 3 × 3 × 3 = (2 × 3) × (2 × 3) × (2 × 3) = (6 × 6 × 6) = 6^{3} Thus, 216 is a perfect cube. And, 6 is the number whose cube is 216. 3. What is the smallest number by which 3087 may be multiplied so that the product is a perfect cube? Solution: Writing 3087 as a product of prime factors, we have: 3087 = 3 × 3 × 7 × 7 × 7 Hence, to make it a perfect cube, it must be multiplied by 3. 4. What is the smallest number by which 392 may be divided so that the quotient is a perfect cube? Solution: Writing 392 as a product of prime factors, we have: 392 = 2 × 2 × 2 × 7 × 7 Clearly, to make it a perfect cube, it must be divided by (7 × 7), i.e., 49. 5. Find the cube of each of the following : (i) (70 ) (ii) 1^{2}/_{3} (iii) 2.5 (iv) 0.06 Solution: (i) (7)^{3} = (7) × (7) × (7) = 343 (ii) (1^{2}/_{3})^{3} = (5/3) ^{3} = 5^{3}/3^{3} = (5 × 5 × 5)/(3 × 3 × 3) = 125/27 (iii) (2.5)^{3} = (25/10)^{3} = (5/2)^{3} = 5^{3}/3^{3} = (5× 5 × 5)/(3× 3× 3) = 125/27 (iv) (0.06) ^{3} = (6/100)^{3} = (3/50)^{3} = 3^{3}/(50)^{3} = (3 × 3 × 3)/(50 × 50 × 50) = 27/125000 Cube and Cube Roots
Cube and Cube Roots  Worksheets


