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CHAPTER 2 CUBES AND CUBE ROOTS 2.1 Introduction Like squares and square roots in Chapter 1, we shall study cubes and cube roots in this Chapter. We first discuss some properties of perfect cubes. We will then discuss patterns of perfect cubes which help us in finding cube roots of small perfect cube numbers. Based on the relationship between units digits of numbers and units digits of their cubes, we discuss a method of finding the cube root of a perfect cube. This method applies to numbers having at the most six digits. The advantage of this method is that there are very few and simple calculations. We then discuss prime factorisation method of finding the cube roots. Like square roots, we do have a division method to find the cube root of a number. However, we do not discuss this method here, as it is a little difficult and beyond the scope of this book. 2.2 Cube of a Number and Perfect Cube Numbers We know that if x is a non-zero number, then x × x × x, written as x 3, is called the cube of x or simply x cubed. Thus, 8 (= 2 × 2 × 2) is the cube of 2 or 2 cubed. Similarly, 27 (= 3 × 3 × 3) is the cube of 3 or 3 cubed. Table 2.1 gives the cubes of the single digits 1 to 9. Table 2.1 : Cubes of the digits 1 to 9 x 1 2 3 4 5 6 7 8 9 x3 1 8 27 64 125 216 343 512 729 Each of the numbers 1, 8, 27, . . ., 729 is the cube of some integer or the other. Such numbers are called perfect cubes (or perfect third powers ). A number n is a perfect cube if there is an integer m such that n=m × m × m. CUBES AND CUBE ROOTS 29 Perfect cubes grow very fast. As m grows from 1 to 9, the perfect cube m3 grows from 1 to 729. Perfect cubes are fairly scattered. Upto 100, there are only four perfect cubes. Upto 1000, there are only ten perfect cubes. (Are you trying to check this fact? Note that 103 = 1000.) How do we examine whether a given number is a perfect cube? If a prime p divides m, then p× p× p will divide m × m × m, i.e., m3. Therefore, if a prime p divides a perfect cube, then p3 also divides this perfect cube. In other words, in the prime factorisation of a perfect cube, every prime occurs three times or a multiple of three times. For example, 64 = 2 × 2 × 2 × 2 × 2 × 2 (= 26) 27000 = 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5 ( = 23 × 33 × 53) Thus, to check whether a number is a perfect cube, we find its prime factorisation and group together triples of the same prime factors. If no factor is left out, the number is a perfect cube. However, if some factor is left as a single factor or a double factor, then the number is not a perfect cube. Example 1 : Examine if (i) 392 and (ii) 106480 are perfect cubes. Solution : (i) 392 = 2 × 2 × 2 × 7 × 7 Here, 7 does not appear in a group of three. Hence, 392 is not a perfect cube. (ii) 106480 = 2 × 2 × 2 × 2 × 5 × 11 × 11 × 11 Here, one prime factor 2 and one prime factor 5 do not appear in a group of three. Therefore, 106480 is not a perfect cube. Example 2 : Examine if 53240 is a perfect cube. If not, find the smallest number by which it must be multiplied so that the product is a perfect cube. Find also the smallest number by which it must be divided so that the quotient is a perfect cube. Solution : 53240 = 2 × 2 × 2 × 5 × 11 × 11 × 11 Here, the prime 5 does not appear in a group of three. Hence, 53240 is not a perfect cube. Further, 5 appears only once. If we multiply the number by 5 × 5, then in the product, 5 will also appear in a group of three and the product will be a perfect cube. Thus, the required smallest number by which the given number should be multiplied is 5 × 5, i.e., 25. Finally, if we divide the given number 53240 by 5, the resulting number has prime factors in a group of three. In fact, 53240 ÷ 5 = 10648 = 2 × 2 × 2 × 11 × 11 × 11. Hence, the required number in this case is 5. 30 MATHEMATICS Remarks 1 : From Table 2.1, we observe that cubes of the digits 1, 4, 5, 6 and 9 are numbers ending in the same digits 1, 4, 5, 6 and 9, respectively. However, 2 and 8 make a pair in the sense that the cube of 2 ends in 8 and the cube of 8 ends in 2. Numbers 3 and 7 make another pair of this kind, i.e., the cube of either ends in the other (33 = 27, 73 = 343). Further, 103 = 1000 shows that if a number ends in a zero, its cube will end in three zeros. These observations will help us in finding the cube roots of perfect cubes. 2. If a number is negative, then its cube is also negative. For example, (–1)3 = (–1) × (–1) × (–1) = –1 = –13 (–2)3 = (–2) × (–2) × (–2) = – 8 = – 23 (–5)3 = –125 = –53, (– m)3 = – m3 This shows that negative numbers may also be perfect cubes. This is in contrast to perfect squares, which are never negative. 2.3 Finding the Cube of a Two-digit Positive Number (Alternative Method) The cube of a number can be obtained by multiplying the number with itself three times. To find x3, we may first find x2 and then x2 × x. Here we shall discuss an alternative method of finding x3, when x is a two digit number. Let x = ab, where a is the tens digit and b is the units digit. Recall that for (ab)2, we formed three columns a2 | 2a × b | b2. Like in the method of finding x2, here too we form columns. We use the Identity (a + b)3 = a3 + 3a2 b + 3ab2 + b3. (You will learn this Identity in Chapter 6.) For (ab)3 we form four columns a3 | 3a2 × b | 3a × b2 | b3 Rest of the procedure is as before, i.e., we retain the units digit after addition and take the remaining digits to the next column for addition.We shall illustrate the method by means of examples. Example 3 : Find 423 using alternative method. Solution : Here, a = 4, b = 2. Thus, the four columns are : a3 | 3a2 × b | 3a × b2 | b3 4 3 = 64 + 10 74 74 ∴ 2 3×4×2 3×4×2 = 96 +4 100 0 423 = 74088 = 48 2 2 3 =8 8 8 CUBES AND CUBE ROOTS 31 Example 4 : Find 873 by alternative method. Solution : Here, a = 8, b = 7. In a situation where a and b are not small, it may not be possible to calculate 3a2 × b and 3a × b2 quickly. In such a situation, we simplify the working as follows : 64 49 64 49 ´ 8 ´ 21 ´ 24 ´ 7 2 2 2 2 a b b a 512 1344 1176 343 a 3b 3a b +146 + 121 + 34 , i.e., 3 2 2 3 1210 658 1465 a 3a × b 3a × b b ∴ 658 873 = 658503 5 0 Example 5 : Find the cubes of the following numbers using alternative method : (i) 27 (ii) 45 (iii) 81 Solution : (i) For 273, we have 49 4 4 6 21 2 294 8 84 +11 +32 + 34 328 19 116 8 19 6 ∴ 49 7 343 3 273 = 19683 (ii) For 453, we have 16 25 25 16 15 12 5 4 64 240 300 125 +27 + 31 + 12 91 271 312 2 5 1 91 ∴ 453 = 91125 3 32 MATHEMATICS (iii) For 813 , we have 64 64 3 8 512 192 +19 +2 531 194 531 ∴ 1 24 24 1 1 1 4 1 4 813 = 531441 EXERCISE 2.1 1. Write the units digit of the cube of each of the following numbers : 31, 109, 388, 833, 4276, 5922, 77774, 44447, 125125125 2. Find the cubes of the following numbers: (i) 35 (ii) 56 (iii) 72 (iv) 402 (v) 650 (vi) 819 3. Find the cubes of the following numbers using alternative method : (i) 35 (ii) 56 (iii) 72. 4. Which of the following numbers are not perfect cubes? (i) 64 (ii) 216 (iii) 243 (iv) 1728 5. For each of the non-perfect cubes in Question 4, find the smallest number by which it must be multiplied so that the product is a perfect cube. 6. For each of the non-perfect cubes in Question 4, find the smallest number by which it must be divided so that the quotient is a perfect cube. 7. By taking three different values of n, verify the truth of the following statements : (i) If n is even, then n3 is also even. (ii) If n is odd, then n3 is also odd. (iii) If n leaves remainder 1 when divided by 3, then n3 also leaves 1 as remainder when divided by 3. (iv) If a natural number n is of the form 3p + 2, then n3 is also a number of the same type. CUBES AND CUBE ROOTS 8. 33 Write true (T) or false (F) for the following statements : (i) 392 is a perfect cube. (ii) 8640 is not a perfect cube. (iii) No perfect cube can end with exactly two zeros. (iv) There is no perfect cube which ends in 4. (v) For an integer a, a3 is always greater than a2. (vi) If a and b are integers such that a2 > b2, then a3 > b3. (vii) If a divides b, then a3 divides b3. (viii) If a2 ends in 9, then a3 ends in 7. (ix) If a2 ends in 5, then a3 ends in 25. (x) If a2 ends in an even number of zeros, then a3 ends in an odd number of zeros. 2.4 Cube Roots If n is a perfect cube, then for some integer m, n = m3. Here, the number m is called the cube root of n. Thus, a number m is a cube root of a number n, if m3 = n. For example : 2 is a cube root of 8, because 23 = 8. 5 is a cube root of 125, because 53 = 125. 11 is a cube root of 1331, because 113 = 1331. If m is a cube root of n, we write m = 3 n. Thus, 2 = 3 8 , 5 = 3 125 and 11 = 3 1331 . In Tables 2.2 and 2.3, we list respectively all the perfect cubes upto 1000 and their cube roots. Table 2.2 Table 2.3 m 1 2 3 4 5 6 7 8 9 10 m3 1 8 27 64 125 216 343 512 729 1000 n 1 8 27 64 125 216 343 512 729 1000 3 n 1 2 3 4 5 6 7 8 9 10 34 MATHEMATICS Remark : We use the symbol ‘3 ’ to denote cube root in the manner we used ‘2 ’ to denote square root. It is a mere convenience that we omit 2 and use ‘ ’ to denote square root. Strictly speaking, we should use the symbol ‘2 ’ for square root. However, we never omit 3 while writing ‘3 ’ for cube roots. We now discuss some methods of finding cube roots of perfect cubes. 2.5 Cube Root Through a Pattern Like squares of natural numbers, cubes too have some interesting patterns: 23 = 8 ∴ 23 – 13 33 = 27 ∴ 33 – 23 43 = 64 ∴ 43 – 33 . . . 93 = 729 ∴ 93 – 83 Also, =7 =1+1×6=1+2× 1× 3 = 19 = 1 + 1 × 6 + 2 × 6 = 1 + 3 × 2 × 3 = 37 = 1 + 1 × 6 + 2 × 6 + 3 × 6 = 1 + 4 × 3 × 3 = 217 = 1 + 1 × 6 + 2 × 6 + . . . + 8 × 6 =1+9× 8× 3 1 = 13 1 + 7 = 23 1 + 7 + 19 = 33 1 + 7 + 19 + 37 = 43 . . . 1 + 7 + 19 + . . . + 217 = 93 Note that 23 is the sum of first 2 numbers of 1, 7, 19, 37, . . . . Similarly, 33, 43, . . ., 9 are the sums of first 3, 4, . . . , 9 numbers respectively out of 1, 7, 19, . . ., 217. These numbers may be obtained by putting n = 1, 2, 3, ... in 1 + n × (n – 1) × 3. Thus, to find the cube root of a perfect cube, we go on subtracting 1( = 1 + 1 × 0 × 3), 7 (= 1 + 2 × 1 × 3), 19 (= 1 + 3 × 2 × 3), 37 (= 1 + 4 × 3 × 3), . . . etc., from the given number till we get a zero. The number of times the subtraction is carried out, gives the cube root. For example, 3 CUBES AND CUBE ROOTS 35 216 – 1 = 215, 215 – 7 = 208, 208 – 19 = 189, 189 – 37 = 152, 152 – 61 = 91, 91 – 91 = 0. Since we have subtracted six times to get 0, therefore 3 216 = 6. This method may be used to find the cube roots of small numbers. This method may also be used to find the smallest number to be added or subtracted to make a non-perfect cube a perfect cube. Example 6 : Examine if 400 is a perfect cube. If not, then find the smallest number that must be subtracted from 400 to obtain a perfect cube. Solution : 400 – 1 = 399, 399 – 7 = 392, 392 – 19 = 373, 373 – 37 = 336, 336 – 61 = 275, 275 – 91 = 184, 184 – 127 = 57 The next number to be subtracted is 169 which is greater than 57. Therefore, the process of successive subtraction does not give zero. Hence, 400 is not a perfect cube. If we subtract 57 from 400, the above process will give zero after 7 successive subtractions. Therefore, 400 – 57 = 73. Thus, 57 is the required number (and 343 is the resulting perfect cube). Similarly, if 112 is added to 400, the sum 512 is a perfect cube. (Why ?) 2.6 Cube Root Using Units Digit We shall now describe a method that can be used to find cube roots of perfect cubes having at the most six digits. By looking at Table 2.2, we observe that the cube of a number ending in 0, 1, 4, 5, 6 and 9 ends in 0, 1, 4, 5, 6 and 9 respectively. However, the cube of a number ending in 2 ends in 8 and vice versa. Similarly, the cube of a number ending in 3 or 7 ends in 7 or 3 respectively. Thus, by looking at the units digit of a perfect cube number, we can determine the units digit of its cube root. Now consider a number which is a perfect cube and has at the most six digits. The cube root of such a number has at the most two digits, because the least seven digit number is 1000000 (= 1003) and its cube root 100 is a three digit number. We determine the two digits of the cube root as follows : Step 1 : Look at the digit at the units place of the perfect cube and determine the digit at the units place in the cube root as discussed above. Step 2 : Strike out from the right, last three (i.e., units, tens and hundreds) digits of the number. If nothing is left, we stop. The digit in Step 1 is the cube root. 36 MATHEMATICS Step 3 : Consider the number left from Step 2. Find the largest single digit number whose cube is less than or equal to this left over number. This is the tens digit of the cube root. Example 7 : Find the cube roots of the following numbers : (i) 512 (ii) 2197 (iii) 117649 (iv) 636056 Solution : (i) 512 : The units digit of 512 is 2. Therefore, the digit at the units place in the cube root is 8. Since no number is left after striking out the units, tens and hundreds digits of the number, the required cube root is 8. (ii) 2197 : Here, units digit is 7. Therefore, units digit of the cube root is 3. After striking out the last three digits from the right, we are left with the number 2. Now 1 is the largest number whose cube is less than 2. Therefore, the tens digit is 1. Thus, the required cube root is 13. (iii) 117649 : Here, units digit is 9. Therefore, the units digit of the cube root is 9. Striking out the last three digits from the right, the number left is 117. Now 43 = 64 < 117 and 53 = 125 > 117. Hence, the tens digit of the cube root is 4. 3 117649 = 49 ∴ (iv) 636056 : Here, units digit of the cube root is 6. (Why ?) Also, 83 < 636 and 93 > 636. Hence, tens digit of the cube root is 8 3 ∴ 636056 = 86 2.7 Cube Root by Prime Factorisation We have already observed that in the prime factorisation of a perfect cube, primes occur in triples. We, therefore, can find 3 n using the following algorithm : 1. Find the prime factorisation of n . 2. Group the factors in triples such that all three factors in each triple are the same. 3. If some prime factors are left ungrouped, the number n is not a perfect cube and the process stops. 4. If no factor is left ungrouped, choose one factor from each group and take their product. The product is the cube root of n. CUBES AND CUBE ROOTS Example 8 : Find the cube root of : (i) 91125 (ii) 531441 (iii) 551368 Solution : (i) 91125 = 5 × 5 × 5 × 3 × 3 × 3 × 3 × 3 × 3 ∴ 3 91125 = 5 ´ 3 ´ 3 = 45 (ii) 531441 = 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 ∴ 3 531441 = 3 ´ 3 ´ 3 ´ 3 = 81 (iii) 551368 = 2 × 2 × 2 × 41 × 41 × 41 ∴ 3 551368 = 2 ´ 41 = 82 37 2 551368 2 275684 2 137842 41 68921 41 1681 41 5 91125 5 18225 5 3645 3 729 3 243 3 81 3 27 3 9 3 3 531441 3 177147 3 59049 3 19683 3 6561 3 2187 3 729 3 243 3 81 3 27 3 9 3 2.8 Cube Roots of Negative Numbers Consider the perfect cubes 27 ( = 33) and 343 (= 73). We know that 27 × 343 = 9261 is also a perfect cube, because a3 × b3 = (a × b)3 holds for integers a and b. Now we find the cube root of 9261 and observe that i.e., 3 27 ´ 343 = 3 27 ´ 3 343 38 MATHEMATICS There is nothing special about 27 or 343. In fact, for any two perfect cubes x and y, we have x´ y = 3 3 x ´ 3 y i.e., cube root of a product of two perfect cubes is the product of their cube roots. Making use of this result, we find the cube roots of negative integers. For a positive integer m –m=–1× m ∴ 3 However, ∴ -m = 3 3 3 -1 ´ 3 m 3 -1 = - 1, because (–1) = – 1 -m = - 3 m Example 9 : Find the cube roots of (i) – 125, (ii) – 343 and (iii) – 2197. Solution : (i) 3 -125 = - 3 125 = – 5 ( Q 53 = 125) (ii) 3 -343 = - 3 343 = – 7 ( Q 73 = 343) (iii) 3 -2197 = - 3 2197 = – 13 ( Q 133 = 2197) 2.9 Cube Roots of Rational Numbers Like the cube root of the product of two perfect cubes, we have the following result for the cube root of the quotient of two perfect cubes : x If x and y (≠ 0) are perfect cubes, then 3 y = 3 3 x , i.e., the cube root of the quotient y of two perfect cubes is the quotient of their cube roots. Note that 3 x and 3 y are integers and 3 y ¹ 0. Hence, 3 3 x is a rational number. y Thus, the cube root of a rational number whose numerator and denominator are perfect cubes, is also a rational number. The numerator of the cube root is the cube root of the numerator of the given number and the denominator of the cube root is the cube root of the denominator of the given number. CUBES AND CUBE ROOTS Example 10 : Find the cube roots of (i) Solution : (i) 3 343 = 125 3 343 = 3 125 3 3 39 -2197 -27 343 , (ii) and (iii) . 1331 512 125 7´7´7 7 = 5 5´5´5 (ii) (iii) 3 - 3 2197 -2197 - 3 13 ´ 13 ´ 13 -13 13 = 3 = 3 = =1331 11 11 1331 11 ´ 11 ´ 11 EXERCISE 2.2 1. Find the cube roots of the following numbers by successive subtraction of numbers 1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, . . . : (i) 64 2. (ii) 512 (iii) 1728 Using the method of Question 1, examine if the following numbers are perfect cubes : (i) 130 (ii) 345 (iii) 792 (iv) 1331 3. Find the smallest number that must be subtracted from those numbers in Question 2 which are not perfect cubes so as to make them perfect cubes. What are the corresponding cube roots? 4. Find the units digits of the cube roots of the following numbers : (i) 226981 (ii) 13824 (iii) 571787 (iv) 175616 5. Find the tens digit of the cube root of each of the numbers in Question 4. 6. Find the cube roots of the following numbers by finding their units and tens digits : (i) 389017 7. (iii) 110592 (iv) 46656 Find the cube roots of the following numbers using prime factorisation : (i) 250047 8. (ii) 91125 (ii) 438976 (iii) 592704 (iv) 614125 Find the cube roots of (i) – 226981 (ii) – 13824 (iii) – 571787 (iv) – 175616 40 MATHEMATICS 9. Find the cube roots of the numbers 2460375, 20346417, 210644875, 57066625 using the fact that : (i) 2460375 = 3375 × 729 (ii) 20346417 = 9261 × 2197 (iii) 210644875 = 42875 × 4913 (iv) 57066625 = 166375 × 343 [Hint : a3 b3 = (ab)3] 10. Find the cube roots of : 343 729 3375 9261 (ii) (iii) (iv) (i) 166375 2197 4913 42875 11. Examine, whether or not each of the following numbers has a cube root. If not, find the smallest number by which the number must be multiplied so that the product has a cube root. (i) 3087 (ii) 33275 (iii) 120393 12. Find the smallest number by which the numbers in Question 11 must be divided so that the quotient has a cube root. Things to Remember 1. A number n is a perfect cube, if there is an integer m such that n = m3. 2. If n is a perfect cube and n = m3, then m is a cube root of n. A cube root of n is written as 3 n. 3. The units digit of the cube root of a perfect cube can be determined with the help of the units digit of the perfect cube. 4. The cube root of a perfect cube can be obtained by prime factorisation of the number. 5. The cube root of a product of two perfect cubes is the product of the cube roots of the perfect cubes, i.e., 3 6. 3 a . 3 b , where a and b are perfect cubes. The cube root of a quotient of two perfect cubes is the quotient of their cube roots, i.e., 3 7. ab = a = b 3 3 a , b ≠ 0, where a and b are perfect cubes. b The cube root of a negative perfect cube is negative.