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3 times tens2 equals 3 × 402 = 4800; dividing by 4800, we find the units to be 5. We then find 3 times tens X units equal to 3 x 40 x 5 = 600, and units2 =52=25, and adding these and multiplying by units we have (3 tens2+3 tens x units + units) X units, which equals 5425 × 5: 27125; subtracting, nothing remains, hence the cube root of 91125 is 45

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OPERATION.

91125(40

403 =64000 5

3 X 4024800 3 X 40 X 5= 600

27125 45

GEOMETRICAL SOLUTION.-Let Fig. 1 represent the cube which contains 91125 cubic units, then our object is to find the number of linear units in its edge. The number of terms in the root, found as before, is two. The greatest number of tens whose cube is contained in the given number is 4 tens. Let A, Fig. 1, represent a cube whose sides are 40, its contents will be 403 64000. Subtracting 64000 from 91125 we find a remainder of 27125 cubic units, which by removing the cube A from Fig. 1, leaves a solid represented by Fig. 2.

52=

251

5425 27125

Inspecting this solid, we perceive that the greater part of it consists of the three rectangular slabs, B, C, and D, each of which is 40 units in length and breadth, hence if we divide 27125 by the sum of the areas of one face of each regarded as a base, we can ascertain their thickness. The area of a face of one slab is 402-1600, and of the three, 3 x 1600, 4800, and dividing 27125 by 4800 we have a quotient of 5, hence the thickness of the slab is 5 units.

=

Removing the rectangular slabs, there remain three other rectangular solids, E, F, G, as shown in Fig. 3, each of which is 40 units long and 5 units thick, hence the surface of a face of each is 40 x 5=200 sqnare units, and of the three it is 3 x 40 x 5600 square units.

Finally removing E, F, and G, there remains only the little corner cube H, Fig. 4, whose sides are 5 units, and the surface of one of its faces 52-25 square units. We now take the sum of the surfaces of the solids remaining after the removal of the cube A, and multiply this by the common thickness, which is 5, and we have their solid contents equal to (4800+ 600 +25) × 5 = 27125 cubic units, which, subtracted from the number of cubic units remaining after the removal of A, leaves no remainder. Hence the cube which contains 91125 cubic units is 40+ 5, or 45 units on a side.

NOTE. This can also be explained by building up the cube instead of separating it into its parts, for which see Manual.

647. We will now solve a problem with three figures in the root, indicating the solution by means of letters, and abbreviating the operation as in practice. A point like a period indicates the multiplication of the letters.

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NOTES.-1. By the geometric method, when there are more than two figures we remove the first cube, rectangular slabs and solids, and small cube, and we have remaining three slabs, three solids, and a small cube, as before. 2. The method employed in actual practice is derived from the other by omitting ciphers, using parts of the number instead of the whole number each time we obtain a figure of the root, etc. It will also be seen that y separating the number into periods of 3 figures each, we have the number of places in the root, the part of the number used in obtaining each figure of the root, etc.

Rule.-I. Begin at units and separate the number into periods of three figures each.

II. Find the greatest number whose cube is contained in the left hand period, write it for the first term of the rool, subtract its cube from the left hand period, and annex the next period to this remainder for a dividend.

III. Multiply the square of the first term of the root by 300 for a TRIAL DIVISOR; divide the dividend by it, and the result will be the second term of the root.

IV. To the trial divisor add 30 times the product of the

econd term of the root by the first term, and also the square of the second term; their sum will be the TRUE DIVISOR. པ Multiply the true divisor by the second term of the rooi · subtract product from the dividend, and annex the nex erind for another dividend. Square the root now found, multiply by 300, and find the third figure as before, and thus continue until all the periods have o ̈ used.

NOTES.-1. If the product of the true divisor by the tea of the root ex cects the dividend, the root must be diminished by a unit.

2. When a dividend will not contain a trial divisor, place a cipher in the 1000 and two ciphers at the right of the trial divisor, bring down the next period, and proceed as before.

3. To find the cube root of a common fraction, extract the cube root of both terms. When these are not perfect cubes, reduce to a decimal al

then extract the root.

4. By cubing 1, .1, .01, etc., we see that the cube of a decimal contains three times as many decimal places as the decimal; hence, to extract the cube root of a decimal, we point off the decimal in periods of three figures each, counting from the decimal point.

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NEW METHOD OF CUBE ROOT.

648. The New Method of extracting cube root is shorte and more convenient than the ordinary method. The abbre viation consists in obtaining the true and trial divisors by a law which enables us to use our previous work.

NOTE. This method seems to have been approximated by several writers, although I have not found any who present it in the form in which it is bere given.

1. Extract the cube root of 14706125.

SOLUTION.-We find the number of figures in the root, and the first term as before. We write 2, the first term of the root, at the left, at the head of Col. 1st; 3 times its square, with two dots annexed, at the head of Col. 2d; its cube under the first period; then subtract and annex

OPERATION.

1ST COL.

2D COL.

14.706-125(245

2

12.. t. d.

8

4

256

6706

54

1456 C. D.

5824

16

1882125

725

1728. . t. d.

3625

176425 C. D.

882125

the next period for a dividend and divide it by the number in Col. 2d, as a trial divisor, for the second term of the root.

We then take 2 times 2, the first term, and write the product 4 in Col. 1st, under the 2, and add; then annex the second term of the root to the 6 in Col. 1st, making 64, and multiply 64 by 4 for a correction, which we write under the trial divisor; and adding the correction to the trial divisor, we have the complete divisor, 1456. We then multiply the complete divisor by 4, subtract the product from the dividend, and annex the next period for a new dividend.

We then square 4, the second figure of the root, write the square under the complete divisor, and add the correction, the complete divisor and the square for the next trial divisor, which we find to be 1728. Dividing by the trial divisor, we find the next term of the root to be 5.

We then take 2 times 4, the second term, write the product 8 under the 64, add it to 64, and annex the third term of the root to the sum, 72, making 725, and then multiply 725 by 5, giving us 3625 for the next correction. We then find the complete divisor, by adding the correction to the trial divisor; multiply the complete divisor by 5, and subtract, and we have no remainder.

NOTE. The correctness of this method may readily be seen by using letters, and following the changes indicated in the solution.

Rule.-I. Separate the number into periods of three fig. ures each; find the greatest number whose cube is contained in the first period, and write it in the root.

II. Write the first term of the root at the head of the 1st Col.; 3 times its square, with two dots annexed, at the head of 2d Col., and its cube under the first period; subtract, and annex the next period to the remainder for a dividend; divide by the number in 2d Column as a TRIAL DIVISOR, and place the quotient as the second term of the root.

III. Add twice the first term of the root to the number in the first column; annex the second term of the root, multiply the result by the second term, and write the product under the trial divisor for a CORRECTION; add the CORREO

TION to the TRIAL DIVISOR, and the result will be the COMPLETE DIVISOR; multiply the COMPLETE DIVISOR by the last term of the root, subtract the product from the dividend, and annex the next period to the result for a new dividend.

IV. Square the last term of the root, and take the sum of this SQUARE, the last COMPLETE DIVISOR, and the last CORRECTION, annexing two dots for a new TRIAL DIVISOR; divide the dividend by this for the next term of the root.

V. Add twice the second term of the root to the last number in the first column; annex the last term of the root to the sum, multiply the result by the last term, and write the product under the last trial divisor for a CORRECTION; add the CORRECTION to the TRIAL DIVISOR, and the result will be the COMPLETE DIVISOR; use this as before, and thus continue until all the periods have been used.

NOTE.-This rule is indicated in the following formula:

1. TRUE DIVISOR = TRIAL DIVISOR + PRODUCT.

2. TRIAL DIVISOR = =PRODUCT + TRUE DIVISOR + SQUARE.

Extract the cube root of

2. 12326391.

Ans. 231.5. 277167808.

3. 34965783.

Ans. 327. 6. 633839.779.

4. 41063625.

Ans. 652.

Ans. 85.9

Ans. 345.7. 4.080659192. Ans. 1.598.

NOTE. For other methods of extracting the cube root see Brooks's Higher Arithmetic.

APPLICATIONS OF CUBE ROOT.

649. The Applications of cube root to problems involving geometrical volumes, such as cubes, parallelopipedons, spheres, etc., are extensive.

650. The Edge of a cube is equal to the cube root of its

contents.

WRITTEN EXERCISES.

1. What are the dimensions of a cubical chest which shall contain 64000 cubic feet? Ans. 40 ft. 2. Required the number of square feet in one face of a cubical block whose con nts are 405224 cu. ft. Ans. 5476. 3. What is the entire surface of a cube whose cubical contents are 91125 cubic feet? Ans. 12150 sq. ft.

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