the remainder 27 affixing the next period, we have for a dividend 27125.

Excluding the two right hand figures, we divide 271 by 48, which is 3 times 16, the square of the root 4 already found, and annex the quotient figure 5 to the root.

To complete the divisor 48, we annex two Os, making 4800, and add 600, which is 3 times 200, the product of the last quotient figure 5 into 40, the previous root 4 with a 0 annexed, and add also 25, the square of 5.

The divisor thus completed is 5425, which, multiplied by 5, equals the dividend.

2. To extract the cube root of 223648543.

108 223'648'543(607

10800!

1080000+12600+49=1092649

216

7648543
7648543

The left hand period is 223, and the greatest cube number it contains is 216, the root of which is 6. Subtracting, and to the remainder 7 affixing the next period, we have for a dividend 7648.

Excluding the two right hand figures, we divide 76 by 108, which is 3 times 36, the square of the root 6 already found. The quotient figure is 0. We annex 0 to the root, and affix the next period, obtaining a new dividend 7648543.

The given number is separated into periods, to determine the number of figures in the root, and also whether one, two, or three figures on the left, correspond to the first figure in the root. (§ 305).

In the 1st example, the first figure 4 in the root, is 4 tens=40; and its cube = 64000, which, subtracted, leaves the remainder 27125.

The given number 91125-4 tens 3 +3×4 tens 2 × the units of the root, +3×4 tens X the units 2, + the units 3. (§ 306).

And since 4 tens 3 has been subtracted, the remainder 27125=3X4 tens 2 the units, +3×4 tens × the units, 2 + the units 3.

In taking 48=3×42 for an incomplete divisor, we omitted two Os in the right of 4 tens 2. Omitting, therefore, the 25 in the corresponding places of the dividend 27125, we say 48 in 271, 5 times.

Regarding now the quotient 5 as the units to be found in the root, the completed divisor 4800+600+25 is 3×4 tens 2+3x4 tens X5 units +5 units 2; and this divisor multiplied by 5=3×4 tens 2 ×5 units, + 3×4 tens X5 units 2, +5 units 3; =27125; and this added to 4 tens 3 makes up the given number.

Hence the given number 91125—(4 tens +5 units)3 =453; (§ 306), hence also, the root is correctly found by the Rule.

Note. The quotient figure- -being found from an incomplete divisor-will often be a unit or two less than the number of times such divisor is contained in those figures of the dividend, which are taken in dividing.

A more convenient method of forming the Divisors, in Extracting the Cube Root.

§ 303. For the first incomplete divisor, take 3 times the square of the first figure in the root; and with this divisor find the second figure, as before.

Complete the divisor by annexing to it two Os, and adding the product of the last figure in the root with 3 times the other part of the root prefixed to it, multiplied by the last figure.

Each succeeding incomplete divisor will be found by adding to the last complete divisor, the product which completed it, and the square of the last figure in the root. The divisors are all completed in the same manner.

EXAMPLE.

To extract the cube root of 95443993.

95'443'993(457

64

48 31443

4800+625-5425 27125

5425+625+25=6075 4318993 607500+9499-616999 4318993

The first incomplete divisor is 48, equal to 3 times 16, the square of 4,-which divisor gives figure in the root.

To complete the divisor 48, we 625, which is the product of 125, 4 prefixed to it), multiplied by 5. is 5425.

the quotient 5, the second

annex to it two Os, and add (that is, 5 with three times The divisor thus completed

For the next incomplete divisor, we add to the last complete divisor the product 625 which completed it, and 25, the square of 5.

Having found the quotient 7, we complete the divisor 6075 by annexing to it two Os, and adding 9499, which is the product of 1357, (that is, 7 with three times 45 prefixed to it), multiplied by 7. The divisor thus completed is 616999.

This method of forming the divisors, involves the same principle, (§ 306,) with the Rule before given.

Thus, the 625 added to complete the first divisor is 125×5 (120+5) X5=120X5+52=3 times 40×5+52.

And the incomplete divisor 6075=4800+625625+25=3 times 402+3 times 40×5+52,+3 times 40×5+52,+52=3 times 402+6 times 40×5+3 times 52=3 times (402+ twice 40×5+52)=3 times 452.

The divisors are, therefore, the same as required by the Rule. (§307).

Cube Root of Decimals, Imperfect Cubes, &c.

§ 309. In extracting the cube root, an integer is separated into periods from right to left. A decimal must be separated into periods of three figures each, from the decimal point towards the right; and one or two Os must be annexed when necessary to complete the last period.

The number of decimal figures in the root, must be the same as the number of periods in the given decimal.

In finding the root of an imperfect cube, (§ 292), a period of Os may be annexed to the last remainder, and the operation continued in this manner to any required exactness; observing that each period thus annexed must be counted as a decim: 1 period belonging to the given number.

A fraction will be an imperfect cube, if either of its terms is an imperfect cube. Its root in such case will be found, most readily, by extracting the root of its equivalent decimal.

A vulgar fraction annexed to an integer may be reduced to a decimal, and the root of the mixed number be then extracted.

The product of three decimal fractions, contains just as many decimal figures as all the three factors. The cube of a decimal fraction has, therefore, three times as many decimal figures as the decimal itself Hence each decimal period must contain three figures; and the number of decimal figures in the root, must equal the number of decimal periods.

§ 310. The solidity or volume of a body, in cubic measure, is equal to the product of its length into its breadth into its height or thickness.

Since a cube has its length, breadth, and thickness, equal to one another, its solidity is equal to the cube of either of its three sides or dimensions.

Hence, the solidity of a cube being given, either of its three dimensions will be found by extracting the cube root of its solidity.

EXERCISES.

25. What must be the length, breadth, or height of a cube, that its volume or solidity may be 1728 cubic feet?

Ans. 12 feet.

26. What must be the length, breadth, or depth, of a cubical box, that its capacity may be 2000 cubic feet?

Ans. 12.598' ft. 27. What must be the depth of a cubical cistern which shall contain 5000 gallons of water? Of one to contain 500 barrels ? Ans. 9.344' ft.; 14.321' ft. 28. What must be the dimensions of a cubical granary which shall contain 3000 bushels of wheat? One to contain 10000 bushels? Ans. 15.513' ft.; 23.173' ft. 29. What must be the dimensions of a cubical cellar whose capacity shall be equal to that of another 30 feet long, 20 feet wide, and 10 feet deep? Ans. 18.469' ft. 30. The solidity of a cubical block of marble is 1331 cubie

feet. What is the length of the block, and the area of its surface? Ans. 11ft.; area 726 sq. ft. 31. A cubical cistern is to be constructed which shall contain 300 barrels of water; the bottom and walls of which are to be plastered with hydraulic lime. of plastering will there be?

How many square yards Ans. 81.056' sq. yd. crib whose capacity shall

32. A farmer wishes to construct a be 2000 bushels,—its breadth and height to be equal, and each of these one half of its length. What must be the length of the crib ? Ans. 21.512' ft.

33. How many square feet of plank will be required to line a reservoir, open at the top, the capacity of which is to be 10000 gallons of water, and the length, breadth, and depth, equal to one another? Ans. 693.017' sq. ft.

34. The capacity of the reservoir being as in the preceding question, how many feet of plank would be required to line it, allowing its length to be double each of its other dimensions? Ans. 698.482' sq. ft.

Extraction of any Root whatever of a Given Number.

Any root whatever of a given number might be extractedby finding the first figure in the root—with this figure, forming an incomplete divisor-with the quotient figure and the preceding part of the root, completing the divisor-and so on, in a manner dependent on the order of the root to be found.

But this method, which is preferable for the square and cube, becomes too complicated when applied to other roots. By dispensing with the complete divisors, the operation may be simplified, and the process of evolution generalized, as follows:

RULE LIX.

§ 311. To extract any required root of a given number.

1. Separate the given number into periods of two, three, or four, &c., figures each-according to the order of the root required from right to left.

2. From the left hand period subtract the greatest power, of the corresponding order, that is contained in it; and set the root of said power for the first figure of the root required.

3. To the remainder affix the first figure of the next period, for a dividend, and,

for the 2d root, divide by 2×1st power of the root already found; for the 3d root, divide by 3×2d power of the root already found; for the 4th root, divide by 4X3d power of the root already found; and so on, annexing the quotient figure to the root.