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Note.-The same rule must be observed for continuing the operation and pointing for decimals, as in the square root.

EXAMPLES.

1. Required the cube root of 436036824287.

436036824287(7583 the root.
313

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3X32 189.-3X32 X7 3X302 )18900=3X302 X7

3X3 ) 441 3X3X72 3X30 4410=3X30X72 As 27 or 27000 is the greatest cube, ite root is 3 or 30, and that part of the cube is exhausted by this extraction. Collect those terms which belong to the same places, and we have 32 X7=63, and 2X32X7=126, and 63+126=3 X 32 X7=189; and 2X3X72 = 294 and 3X 72 =147, and 294+147=44133X3X72 for a dividend, which divided by the divisor, formed according to the rule, the quotient is 7, for the next figure in the root.

And it is evident, on inspecting the work, that that part of the cube not exhausted is composed of the several products which form the subtrahend, according to the rule.The same may be shown in any other case, and the universality of the rule hence inferred.

The other method of illustration, employed in the square root, is equally applicable in this case. 37=30+7, and 30+72=302 +-2X 30X7+72

30 +7 the multiplier.

303-+-2X302 X7+30X72

302 X7+2X30X72+73

373350653303+3X302 X7+3X30X72 +73 (30+7–37

303

Divisor 3X302 +3X30 18X302 X7+3X30X72+73 div.

3X302X77-3X30X72+73 subtrahend. It is evident that 303 is the greatest cube. When its root is extracted, the next three terms constitute the dividend; and the several products formed by means of the quotient or second figure in the root, are precisely equal to the remaining parts of the powes whose root was to be found.

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758x758X300 =172369200 = 3d Triple square. 758x30

22740 = 3d Triple quotient.

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DEMONSTRATION OF THE REASON AND NATURE OF THE RULE.

A block of wood, or any solid body, having six equal sides, all exactly square, is a CUBE ; the root of which is the measure in length of one of its sides. To gain a distinct understanding of the subject, let the scholar provide himself with little blocks of wood, and build them up into a cubick form, according to the rule.

First make a cubick block of any given size, and mark it with the letter A. Then make three other blocks of a square form, of an indefinite thickness, but all equal to each other, each of which will just cover one side of the block A, and mark them B, C and D. Place these blocks on three adjoining sides of the block A, when there will be deficiencies at the three points where the blocks B, C and D meet, These deficiencies must be filled with three other blocks, each of which must be just equal in length to one side of the block A, and mark these blocks with E, F and G. When the blocks E, F, G are put in their places, there will be a deficiency at the place where The ends of these blocks meet. This deficiency must be filled with another block, which mark H. To illustrate the rule, take the following number.

10648(2
8

a

2 In the first place I seek the greatest cube in the left band period, and place its root, 2, in the quotient. The cube of 2 is 8, which I place under the left hand period, and subtract it thererom, which leaves a remainder of 2. Now as there are two periods in the given number, there must be 2 figures in the root, consequently, 2, in the quotient, does not express 2, merely, but 20 ; and the cube of 20 s 8000, which 8, under the period 10, represents ; thus 8000 of the parts of 10648, are disposed of into a cubick body, the length of each side of which is equal to 20 of those parts, and to render the explanation more plain, we will consider these parts as cubick feet, so that each side of this body is 20 feet square, and this body we will have represented by the block A. Now as each side of this block is 20 feet square, there are 400 feet on each side of it. Now 8000 feet are disposed of in this block, consequently, there are 2648 cubick seet to be added to the block A, in such a manner, that its cubick form will be preserved. To do this, the additions must be made to three sides of the block, and these additions are represented by the blocks B, C and D, each of which containing 400 feet, the sum of the whole is 1200. Thus it is evident, that if there were 1200 feet more, there would be just enough to cover three sides of the block A ; and it is to find the contentş of these three sides, that the rule directs to multiply the square of the quotient by 300." of the quotient shows the superficial contents of one side of the block A. viz. 400, for 2 in the quotient is in reality 20, and 20x20=400, and

, it is because the cipher is not andexed to the quotient figure, that we ,

The square

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182

are directed to multiply the square of the quotient by 300 instead of 3, as in the following work.

10648(2 4

1200 60 8

2 300

4

240

1200 triple square.

1260)2648

2400
2400
240

8 cube of the root.

2.
30

2648 subtrahend.

60 triple quotient. 1200

1260 divisor. The rule directs to “ multiply the quotient by 30." This is to obtain the contents of the blocks E, F and G, and the sum of these is taken for a divisor, because the number of times the dividend contains the divisor, will be equal to the number of feet, the additions to the block, A. are in thickness.

The rule next directs to “ multiply the triple square by the last quotient figure." Now the triple square represents the superficial contents of the three blocks B, C and D, and the last quotient figure shows the thickness of those blocks, consequently, multiplying the triple square by the last quotient Sgure, gives the cubick contents of those blocks. Then the rule directs to multiply the square of the last quotient figure by the triple quotient." Now the triple quotient is the length of the blocks E, F and G, and the quotient figure shows their

breadth, and their thickness; hence, multiplving the square of the last quotient figure by the triple quotient, gives Nue cubick contents of these blocks. Next we are directed to cube figure. This cube shows the cubick contents of the block sum of these is equal to the cubick contents of the blocks B,

and D; and E, F and G.

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APPLICATION AND USE OF THE CUBE ROOT.

PROBLEM 1. To find two mean proportionals between any two given numbers.

Rule 1. Divide the greater number by the less, and extract the cube root of the quotient.

2. Multiply the root, so found, by the least of the given numbers, and the product will be the less.

3. Multiply this product by the same root, and it will give the greater.

EXAMPLES

3

1. What are the two mean proportionals between 6 and 750 ?

750—6=125, and ✓ 125–5. Then 5X6=30 the less, and 30X 5=150 the greater.

Ans. 30 and 150.

2.

What are the two mean proportionals between 56 and 12096 ?

Ans. 336 and 2016.

PROBLEM 2. The side of a cube being given, to find the side of a cube which shall be any nurber of times greater or less than the given cube. *

Rule.-Cube the given side, and if the required cube be greater than the given one, multiply the cube of the given side by the given proportion, and the cube root of the product will be the side of the cube required. But if the required cube be less, divide the cube of the given side by the given proportion, and the cube root of the

quotient will be the side of the required cube. ck s B,

EXAMPLES.

Ans. 39

3

There is a cubick box whose side is 18 inches ; what is the side box that will contain 8 times as much ?

X18-5832, and 5832X8=46656. Then 46656-36 in. Ans. Ans. 4:39.here is a cubick box whose side is 24 inches ; what is the side that will contain one sixty-fourth part as much ?

Ans. 6 inches.

ans. -20527.

lid, called a cube, has its length and breadth and height all equal. As the

solid feet, inches, &c. in a cube are found by multiplying the height and Ans.breadth together ; that is, by multiplying one side into itself twice, the third

a number is called the cube of that number. The solid contents of similar
in proportion to each other, as the cubes of their similar sides or diameters.

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