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The same may be shown for the cube of a number consisting of any number of figures. Hence the

Principle. The cube of a number contains three times as many figures as the number itself, or three times as many less one, or three times as many less two.

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whose cube is contained in 15625 is 2 tens, or 20 units. This is shown in Fig. 1 by the cube A, whose sides are 20 units and whose volume is 203, or 8000. Subtracting 8000 from 15625, the remainder is 7625. This we see by Fig. 2 to consist principally of three equal square solids, B, C, and D, each of which is 20 units square. Since they nearly complete the cube, their volume is nearly 7625. If we divide 7625 by their surface, we shall find their thickness. The surface of each square solid is 202, and that of the three solids is 3 × 202 1200 square units. Dividing 7625 by 1200, we have a quotient of 5 units; hence the thickness of these square solids is 5 units.


Removing these three square solids, there are three rectangular solids remaining, E, F, G (Fig. 3), each being 20 units long and 5 units thick; hence the surface of a face is 20 x 5 100 square units, and that of the three is 300 square units.


There remains a small cube, H (Fig. 4), whose sides are 5 units, and the surface of one face of which is 52, or 25 square units.

Taking the sum of these three surfaces, we have 1525 square units, and multiplying this by the thickness, 5, we have 7625 cubic units; subtracting, there is no remainder. Hence the cube root of 15625 is 20 + 5, or 25.

Omitting the ciphers, the process will stand as follows:

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From this example we derive the following


Begin at the unit's place and separate the number into periods of three figures each.

Find the largest number whose cube is contained in the lefthand period; write it as the first figure of the root; subtract its cube from the left-hand period, and annex the next period to the remainder.

Take three times the square of the root found, for a Trial Divisor; divide the remainder, omitting the last two figures, by this divisor, and annex the quotient to the root.

Add to the trial divisor three times the product of the first and second terms of the root, written one figure to the right, and also the square of the second term of the root, written one figure to the right; their sum will be the True Divisor.

Multiply the complete divisor by the second term of the root, and subtract the product from the remainder.

If other periods remain, take three times the square of the root already found for the next trial divisor, and proceed as before.

When a cipher occurs in the root, annex two ciphers to the trial divisor and bring down the next period.

3. Find the cube root of 41063625.

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229. The following Short Method of cube root is the shortest and most convenient method yet used. The abbreviation consists in obtaining the successive trial divisors by using the previous work. It is readily explained either by the blocks or by the algebraic formula.

19. Find the cube root of 48228544.

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It is not necessary to give a full rule for this method, as it is merely a modification of the previous method. It will be easily remembered by means of the following statement:

Find the first trial divisor by the usual method.

The second trial divisor is obtained by adding to the first complete divisor the two corrections required to form it, together with the square of the second term.

This method holds good for any trial divisor, and will be found to save a great deal of labor in extracting the cube root.

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We thus see that the cube of a decimal contains three times as many decimal places as the decimal itself; hence to extract the cube root of a decimal we point off the decimal into periods of three figures each, beginning at the decimal point. The number of periods to the right of the decimal point shows the number of places in the decimal portion of the root.

The method of ART. 229 may be used to find the cube root of a decimal if the decimal point is inserted in its proper place in the root.

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