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therefore, the required root is composed of 2 tens, plus a certain number of units less than ten.

We now subtract 8, the cube of the tens, from 13, and bring down the next period, 824. We have now 5824, which contains the three remaining parts of the cube, viz: Three times the product of the square of the tens into the units, plus three times the product of the tens into the square of the units, plus the cube of the units. Now, as the square of tens gives hundreds, it follows, that three times the square of the tens into units must be contained in 58, which we separate from 24 by a line. If we now divide 58 by three times the square of the tens, we shall obtain the units of the required root. We may ascertain whether the unit figure be right, by cubing the quotient, or by applying the following principle: The difference between the cubes of two consecutive numbers is equal to three times the square of the least number, plus three times this number, plus 1. Thus, the difference between the cube of 3 and the cube of 4, is equal to 9×3+3×3+1=37, which is the difference between the cube of 3 and the cube of 4. Therefore, had we written 3 in the unit's place, the remainder would have been equal to 3 times the square of 23, plus three times 23, plus 1, which would show that the unit figure must be increased.

Thus far the illustration has been general,―applied to numbers merely-numbers in the abstract. We may now apply it to solid bodies. Numbers which represent, or stand for things, are called concrete, as question first below.

EXAMPLES.

Art. 231.-1. What is the length of one side of a solid block containing 13824 solid inches, or what is the cube root of 13824 ?

OBS.-The foregoing operation can be better understood by blocks prepared for the purpose. It is necessary to have one cubical block, of a convenient size, to represent the greatest cube in the left-hand period, and three other blocks, equal to the sides of the first block, but of indefinite thickness, to represent the additions upon the sides. Then three other blocks, equal in length to the sides of the cube, and their other dimensions equal to the thickness of the additions on the sides of the cube. Lastly: a small cubic block, of dimensions equal to the thickness of the additions, to fill the deficiency at the corner. By placing these blocks as above described, the several steps in the operation may be easily understood. It may be observed, however, that this illustration would serve only for concrete numbers, as in the above question.

Having distinguished the given number into periods of three figures each, denoted by the index of the root, we perceive, by the number of periods, that the root will consist of two figures. As the cube of ten cannot be less than a thousand, 10 x 10 x 10=1000, we look for the cube of tens in the second, or left

Operation.

23=2x2x2=8

22 x 300+601260) 5824

hand period. We find, by trial, the greatest cube in 13, or 13000, to be 8,

13824(24 root. or 8000, and its root, 2 or 2 tens, (the length of one side of the cube, Fig. 4,) which we place in the quotient, as the first figure of the root, and its cube, 20 x 20×20=8000, under that period; and, of 5, or 5000-to which Had the cube contained but

1200 × 44800 60 × 4X4 960 4X4 X4 64 5824

subtracting it, we have a remainder
we bring down the next period.
8000 solid inches, we should now
have found its root, or the length
of one side. But we have 5824
inches to be added to the cube,
and in such a manner that its
cubic form shall not be altered.
It is obvious, that an equal ad-
dition must be made on three
sides. As each side is 20 inches
square, we have 20 x 20 x 3=
1200; or, which is the same thing,

90

[subsumed][merged small][merged small][graphic]

Fig. 5.

20

20

multiply the square of the quotient by 300. 2x2 x300= 1200 inches surface, to which the additions are to be made. It will be seen (Fig. 5) that there are three deficiencies along the sides, a a a, where the additions meet, 20 inches in length, 20 x 3 =60, or multiply the quotient by 30; 2 × 30=60. We have, then, 1200+60=1260, which may be considered the points where & the additions are to be made. Then 58241260-4 inches, the thickness of the addition, or the

[graphic]

20

[blocks in formation]

second figure of the root.

Then the

Fig. 6. 20

The area of the sides multiplied by the thickness, 1200 x 4-4800 inches, the amount of the addition upon the sides. number of inches necessary to fill the deficiencies where the additions on the sides meet, is 60 x 4 X4 960 inches. Still there is → a deficiency of a small cube in the corner, (Fig. 6,) whose dimensions are equal to the thickness of the additions: 4 x 4X4= 64 inches. This supplied, and the cube is completed. (Fig. 7.) The sum of all the additions will

[graphic]

90

90

be a subtrahend equal to the dividend; 4800+960+64=5824. We have now found the length of one side of the cube to be 24 inches. Proof by Involution :

24 x 24 X 24=13824.

Art. 232.-Hence it appears, that a cube is a solid body, having six equal sides, and its cube root is the length of one of those sides.

From the foregoing example and illustration we derive the following

RULE.

24

Fig. 7.

[merged small][graphic][subsumed]

I. Distinguish the given number into periods of three figures each, beginning at the right hand.

II. Find the greatest cube in the left-hand period, and place its root as a quotient in division.

III. Subtract the cube from said period, and to the remainder bring down the next period, for a dividend.

IV. Multiply the square of the quotient by 300, calling it the triple square, and the quotient by 30, calling it the triple quotient, and the sum of these call the divisor.

OBS.-The triple quotient is not indispensable in forming the divisor.

V. Seek how many times the divisor is contained in the dividend, and place the result in the quotient, for the second figure of the root.

VI. Multiply the triple square by the last quotient figure, and write the product under the dividend; multiply the triple quotient by the square of the last quotient figure, and place this product under the last; under these write the cube of the last quotient figure, and call their sum the subtrahend. Subtract the subtrahend from the dividend, and to the remainder bring down the next period, for a new dividend, and proceed as before, till the work is finished.

[blocks in formation]

3. What is the cube root of 941192? 4. What is the cube root of 6331625 ? 5. What is the cube root of 11543176000? 6. What is the cube root of 34.328125 ? 7. What is the cube root of .000729 ? 8. What is the cube root of .003375 ? 9. What is the cube root of 5? of 3? 10. What is the cube root of 8 ? 11. What is the cube root of 216?

125

343

[blocks in formation]

Ans. 98.

Ans. 185.

Ans. 2260.
Ans. 3.25.
Ans. .09.
Ans. .15.

Ans. .

Ans. .

Ans. 13.

Ans..

14. A certain hill contains 11543176 cubical feet.

What

is the length of one side of a cubical mound, containing an equal

number of feet?

Ans. 226 feet.

15. The contents of an oblong cellar is 9261 cubical feet. What is the length of one side of a cubical cellar, of the same capacity? Ans. 21 feet.

16. A merchant bought cloth to the amount of $393.04, but forgets the number of pieces, and also the number of yards in each piece, and what the cloth cost per yard; but remembers that he paid as many cents per yard as there were yards in each piece, and that there were as many in each piece as there were pieces. What did he pay per yard? Ans. 34 cents. 17. What is the width of a cubical vessel, containing 75 wine gallons, each 231 cubic inches?

18. Required the side of a cubic box that shall contain a bushel ? Ans. 12.9-inches.

Art. 233.-Solids of the same form are to one another as the cubes of their similar sides, or diameters.

EXAMPLES.

1. If a bullet, weighing 72 lbs., be 8 inches in diameter, what is the diameter of a bullet weighing 9 lbs. ?

[blocks in formation]

2. A bullet, 2 inches in diameter, weighs 4 lbs. What is the weight of a bullet 5 inches in diameter ? Ans. 62 lbs. 3. If a silver ball, 9 inches in diameter, be worth $400, what is the worth of another ball, 12 inches in diameter ?

Ans. $948.148+.

Art. 234.-To find two mean proportionals between two numbers.

RULE.

Divide the greater by the less, and extract the cube root of the quotient: multiply the lesser number by this root, and the product will be the lesser mean; multiply this mean by the same root, and the product will be the greater mean.

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