WRITTEN EXERCISE Find the cube root of each of the following: 3. 9261. 4. 5832. 7. 29,791. 8. 46,656. 11. 226,981. 12. 300,763. 1. 4913. 2. 6859. 5. 13,824. 6. 19,683. 9. 132,651. 10. 157,464. 13. 551,368. 14. 753,571. 17. 592,704. 18. 778,688. 19. 857,375. 20. 970,299. 15. 884,736. 16. 941,192. By the method of § 587, find the cube of each of the following, proving the result by multiplication: 588. Number of figures in the cube root. It is easy to tell in advance the number of figures in the cube root of a perfect cube. For The cube of units has 3, 2, or 1 figure, since 93729, 43 = 64, 13 = 1. The cube of a number of 2 integral places has 6, 5, or 4 integral places, since The cube of a number of 3 integral places has 9, 8, or 7 integral places, and so on. Therefore 589. If a cube number be separated into periods of three figures each, beginning at the decimal point, the number of periods will equal the number of figures in the root. Thus, V1,771,561 has 3 integral places; 590. Cube root by the formula for (f+n)3. 1. Find √2197. If we let and then f = the found part of the root, n = the next figure of the root, (f + n)3 = ƒ3 + 3 ƒ2n + 3 ƒn2 + n3. (§ 587) Therefore if we take away ƒ3 (B on page 504), we shall have 3 ƒ2n + 3 ƒn2 + n3 (D, page 504, where each square block is ƒ2n, each oblong ƒn2, and the small cube is n3). If we divide this by 3ƒ2, we shall find approximately n (as in E, page 504, where we divided the number of units of volume by the number of units of area of the three squares). The greatest cube of tens in 2197 is 1000, for 203 = 8000, and this is greater than 2197. Therefore ƒ3 is 1000. Since ƒ3 = 1000, f is 10. The remainder, 1197, contains 3 ƒ2n + 3 ƒn2 + n3, because from a number containing (ƒ+ n)3, or ƒ3 + 3 ƒ2n + 3 ƒn2 + n3, ƒ3 has been subtracted. Dividing this by 3 ƒ2 (as seen also on page 504), we approximate n, and n = 3. Now the quantity which multiplied by n equals 3 ƒ2n + 3 ƒn2 + n3, and thus, with the f3 already found, completes the cube of f+n, is 3 f2 + 3 fn + n2. (See also page 504.) Therefore we add to 3 f2 the quantity that will make this, that is, 3 fn + n2. Now 3 fn + n2 = 3 × 10 × 3 + 32 = 99, and 30099 399. Multiplying 399 by 3, the result is 1197, exactly completing the cube. It is easily seen that 2197 must be the cube of 13, for (§ 587) 133 = 103 + 3 × 102 × 3 + 3 × 10 × 32 + 33 = 1000 + 900 + 270 +9 = = 1000 + 1179, which are exactly the numbers above subtracted from 2197, leaving no remainder. There cannot be any thousands in the root, for 10003 = 1,000,000,000, and this is larger than the given cube. The greatest cube of hundreds in 1,771,561 is 1,000,000, because 2003 is larger than the given number. Therefore ƒ = 100. Then the remainder, 771,561, contains 3 ƒ2n + 3 ƒn2 + n3, because ƒ3, or 1,000,000, has been subtracted from a number containing f3+3 ƒ2n + 3 ƒn2 + n3. Because this contains approximately 3 f2n, if we divide by 3 ƒ2 (or 30,000) we shall find approximately n. We therefore find that n = = 20. Adding 3 fn + n2, or 3 × 100 × 20 + 202, to 3f2, we have 36,400 (as explained on page 507). This multiplied by 20 (that is, by n) equals 728,000, thus completing the cube of 120. That 1,000,000 and 728,000 are together the cube of 120 is evident from the fact that 12031003 + 3 × 1002 × 20 + 3 × 100 × 202 +203 (§ 587)=1,000,000 + 600,000 + 120,000 + 8000 = 1,000,000 + 728,000. We may now consider 120 as the found part of the root, and proceed as before, finding n = 1. The root is therefore 121. The annexed arrangement shows all of the necessary figures, and is the one recommended after the process is understood. It simply preserves the figures actually used from step to step, not writing the others. In case there are more figures in the root we may subtract and proceed as before, dividing again by three times the cube of the found part. 591. The following are therefore seen to be the steps in extracting cube root: 1. The number being already separated into three-figure periods, beginning at the decimal point, find the greatest cube in the left-hand period. This is the cube of the number represented by the first figure. 2. Subtract this cube, bringing down but one period. 3. Divide this remainder by three times the square of the found part considered as tens (the so-called trial divisor) to find the next figure. 4. To three times the product of the two parts add the square of the second, considered as units, and add all this to the trial divisor, thus making the complete divisor. 5. Multiply the complete divisor by the second, thus completing the cube of the first two. 6. Subtract this, bring down the next period, and proceed as before. The subject of cube root being now so commonly postponed until algebra is studied, a more extended explanation is not felt to be necessary. WRITTEN EXERCISE Extract the cube root of each of the following: MENSURATION 592. Ratio of circumference to diameter. By measuring several circles, dividing each circumference by its diameter, and taking the average of the results, the circumference will be found to be about 34 times the diameter. 593. Value of π. It is proved in Geometry that this ratio of circumference to diameter is more nearly 3.1416. The ratio is denoted in mathematics by the Greek letter π (pi). 594. Formula for circumference. Therefore, if c = circumference, d diameter, and r = radius, we have = = 2 × 34 × radius (nearly). 595. Illustrative problems. 1. Required the circumference when the radius is 7 in. c = 2πr = 2 × 3 More exactly, 2 x 3.1416 x 7 in. = × 7 in. = 44 in. = 2. Required the diameter when the circumference is 2827.44 in. Using π = =3122, state the circumferences of circles of diameters as follows: |