ANALYSIS.-Since 405224 consists of two periods, its cube root will consist of two figures (818, PRIN. 4). Since 224 cannot be a part of the cube of the tens of the root (818, PRIN. 2), the first figure of the root must be found from the first period, 405. The greatest number of tens whose cube is contained in 405000 is 7. Subtracting the cube of 7 tens from the given number, the remainder is 62224. This remainder is equal to the product of three times the square of the tens of the root by the units, plus three times the product of the tens by the square of the units, plus the cube of the units (804, PRIN.) But the product of the square of tens by units cannot be of a lower order than hundreds (818, PRIN. 2); hence the number represented by the last two figures, 24, cannot be a part of three times the product of the square of the tens of the root by the units; the triple product must therefore be found in the part 62200. Hence, if 62200 be divided by 3 × 702, the quotient, which is 4, will be the units' figure of the root or a figure greater than the units' figure. Subtracting 743 from the given number, the result is 0; hence 74 is the required root.

Instead of cubing 74, the parts which make up the remainder 62224 may be formed and added thus:

Or, since 4 is a common factor in the three parts which make up the remainder, these parts may be combined thus:

1. In this example, 14700 is a partial or trial divisor, and 15556 is a complete divisor.

2. If the cube root contains more than two figures, it may be found by a similar process, as in the following example, where it will be seen that the partial divisor at each step is equal to three times the square of that part of the root already found.

RULE.-I. Separate the given number into periods of three figures each, beginning at the units' place.

II. Find the greatest number whose cube is contained in the period on the left; this will be the first figure in the root. Subtract the cube of this figure from the period on the left, and to the remainder annex the next period to form a dividend.

III. Divide this dividend by the partial divisor, which is 3 times the square of the root already found, considered as tens; the quotient is the second figure of the root.

IV. To the partial divisor add 3 times the product of the second figure of the root by the first considered as tens, also the square of the second figure, the result will be the complete divisor.

V. Multiply the complete divisor by the second figure of the root and subtract the product from the dividend.

VI. If there are more periods to be brought down, proceed as before, using the part of the root already found, the same as the first figure in the previous process.

1. If a cipher occur in the root, annex two ciphers to the trial divisor, and another period to the dividend; then proceed as before, annexing both cipher and trial figure to the root.

2. If there is a remainder after the root of the last period is found, annex periods of ciphers and proceed as before. The figures of the root thus obtained will be decimals.

What is the cube root

3. Of 15625 ?

5. Of 1030301 ? 7. 4. Of 166375? 6. Of 4492125 ? 8.

9. Find the cube root of

OPERATION.-/

=

3

8

√/27

Of 1045678375 ?

Of 4080659192 ?

RULE.-The cube root of a fraction may be found by extracting the cube root of the numerator and denominator.

In extracting the cube root of decimal numbers, begin at the units' place and proceed both toward the left and the right, to separate into periods of three figures each.

Extract the cube root

10. Of 1889:

11. Of 1.

12. Of 24.

13. Of 39304.

14. Of .091125.

15. Of 12.812904.

16. What is the cube root of 98867482624 ?

17. What is the cube root of .000529475129 ?

18. Find the cube root of correct to 4 decimal places.

Find the second member of the following equations:

19. 1.44+ 2.55 = ? | 21. 20. VVTH? 22.

.4096 — .2368 = ?

54.872-(21.952)=?

23. 24.8103.823 × (.125) = ?

24. 166 ÷ 64 — (4 × √.512) 64-(4

= ?

GEOMETRICAL EXPLANATION OF CUBE ROOT.

820. What is the length of the edge of a cube whose volume is 15625 cubic feet?

will consist of two figures. The greatest number of tens whose cube is contained in 15000 is 2. Hence, the length of the edge of the cube is 20 feet plus the units' figure of the root. Removing the cube whose edge is 20 feet and whose volume is 8000 cubic feet, there remains a solid whose volume is 7625 cubic feet (Fig. 2). This remainder consists of solids similar to those marked B, C, and D, in Fig. 1 and Fig. 2 of Art. 804.

The volume of a rectangular solid is equal to the product of the area of its base by its height or thickness (472); hence, if the volume be divided by the area of the base the quotient will be the thickness. Now, since the three equal rectangular solids, each of which is 20 feet square and whose thickness is the units' figure of the root,. contain the greater por

X

tion of the 7625 cubic feet, 3 x 202 or 300 × 22 may be used as a trial divisor to find the thickness. Dividing 7625 by 1200 the quotient is 6. But this quotient is too large, for if 6 feet is the thickness, the volume of Fig. 2 will be 3 × 202 × 6+3 × 20 × 62+63, or 9576 cubic feet. Taking 5 feet for the thickness, the volume of Fig. 2 is 7625 cubic feet, for 3 x 20 x 5+ 3 x 20 x 52 +53 (300 × 22 +30 × 2 x5+52) 5=1525 x 5=7625. Hence, 25 feet is the length of the edge of a cube whose volume is 15625 cubic feet.

PROBLEMS.

821. 1. What is the length of the edge of a cubical box that contains 46656 cu. inches?

2. What must be the length of the edge of a cubical bin that shall contain the same volume as one that is 16 ft. long, 8 ft. wide, and 4 ft. deep?

3. What are the dimensions of a cube that has the same volume as a box 2 ft. 8 in. long, 2 ft. 3 in. wide, and 1 ft. 4 in. deep?

4. How many square feet in the surface of a cube whose volume is 91125 cubic feet?

5. What is the length of the inner edge of a cubical bin that contains 150 bushels ?