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If block No. 1 be placed upon a table, the three blocks No. 2, laid against three of its sides, the three blocks No. 3, in the deficiencies left by No. 2, and No. 4 in the corner still unfilled, we shall have a new cube, which will represent THE CUBE OF THE TENS, plus THREE TIMES THE SQUARE OF THE TENS MULTIPLIED BY THE UNITS, plus THREE TIMES THE SQUARE OF THE UNITS MULTIPLIED BY THE TENS, plus THE CUBE OF THE TENS.

This set of blocks is readily applicable in showing the difference between 303 and 373. But it will be easily perceived that the same result would have been obtained, if, instead of 3 inches, and of an inch, we had employed 4 inches, and of an inch, -11 inches, and of an inch,- -or any other numbers to represent tens and units. Therefore, the cube of any number whateveris equal to the cube of the tens+three times the square of the tens the units, three times the tens the square of the units,+the cube of the units.

We will now cube 37, both by our rule and by direct multiplication.

thous.

hund.

441 tens,
343 units,

The cube of the tens,

3x the square of the tens

the

units,

3× the tens the square of the

units,

The cube of the units,

=27

=189

tr. div. 3x 32 = 27,49 3×30×7 6 30

=

=

23653
23653

The cube of 37,

=50653

Let it now be required to extract the cube root of 50653.

50653(37

27

37 37

372 1369

37

373=-50653

Dividing the number into periods, we find there will be two figures in the root. 27 is the greatest cube contained in the first period, and its root 3, is evidently the tens' figure of the root sought. Subtracting 27

comp. div. 3379

from 50, and annexing the next period to the remainder, we have 23653, which must evidently contain 3× the square of the tens the units,+3x the tens the square of the units+the cube of the units, which is the same as, the units X (3× the square of the tens +3× the tens the units+the square of the units). To discover the units' figure, we will employ 3x the square of the tens, or 27 hundreds, for a trial divisor. 27 hundreds is contained in 236 hundreds 8 times. But if we complete our divisor with 8, it will be found too large; we therefore try 7. 3× the tens the units

63 tens or 630. The square of the units is 49 units, which is annexed to the 27 hundreds. Adding these numbers, we obtain 3379 for our complete divisor, which being multiplied by 7, gives

23653,-the remaining part of the root. following

Hence we derive the

RULE.

Separate the number into periods of three figures, by placing a point over the units' figure, and one over each third figure to the left, (and also to the right, if decimals are desired in the root). Write in the quotient the root of the greatest cube contained in the left hand period, and subtract the cube from the period, annexing the second period to the remainder.

Take three times the square of the root already found, for a trial divisor, and find how many times this divisor is contained in the hundreds of the dividend. Place the result in the quotient, and its square at the right of the trial divisor (supplying the place of tens with a zero, if the square is less than ten).

Multiply 30 times the root already found, by the last root figure; this product added to the divisor, will give the complete divisor.

Multiply the complete divisor by the last root figure, subtract the product from the dividend, annex the next period to the remainder, and proceed as before until the whole root is extracted.

If any complete divisor is not contained in its dividend, a zero must be annexed to the root, and two to the trial divisor, and the next period brought down for a new dividend.

If any figure obtained for the root is too great, diminish it by 1 or more, and repeat the work.

[It may be well to illustrate to the pupil the algebraical formation of the cube, by raising 30+7 to the third power.

In forming the complete divisor, we say, "Multiply 30 times the root already formed, by the last root figure," because 3× the tens the units would give tens, and a zero must be annexed for units.

The square of the units is written at the right of the trial divisor, to render the work more compact. This square being units, and the trial divisor hundreds, the two may very properly be united.]

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The trial divisors, after the first, may be more conve niently found, by adding to the last complete divisor the last number used to complete it, and twice the square of the last root figure. Thus in the foregoing example, our third trial divisor= 1096281 16200 162

2× 92

trial divisor, 1112643

1. What is the cube root of 328509? 2. What is the cube root of 1986091? 3. What is the cube root of 9129329 ?

512

4. What is the cube root of 345.357380096?
5. What is the cube root of .062570773 ?
6. What is the cube root of; 25; 100?
7. What is the cube root of 5; 113; 274§?
8. What is the cube root of 6; ; 21; 95; 8.71?

9. What is the cube root of 148877 ? 10. What is the cube root of .0081 ?

1030301

11. What is the cube root of 4.160075243787 ? The solid contents of similar bodies are to each other as the cubes of their diameters, or of their similar sides.

The solid contents of a sphere may be found by multiplying the cube of the diameter by .5236.

12. What are the solid contents of the earth, supposing it a perfect sphere, whose diameter is 7920 miles?

13. If a ball 2 inches in diameter, weighs 11⁄2 pounds, what would be the weight of a similar ball 6 inches in diameter ?

14. What is the side of a cubical box that will hold 1 bushel?

15. What is the side of a cubical pile that contains 258 cords of wood?

16. If a tree 1 foot in diameter, yields 2 cords of wood, how much wood is there in a similar tree that is 3ft. 6in. in diameter ?

17. If a pound avoirdupois of gold is worth $200, and a cubic inch weighs 114oz., what would be the value of a gold ball 1 foot in diameter ?

18. What is the size of a ball that weighs 27 times as much as one 3 feet 6 inches in diameter ?

19. If a hollow sphere 3 feet in diameter and 21 inches thick, weigh 12 tons, what would be the dimensions of a similar sphere that would weigh 324 tons?

20. What is the side of a cubical block of wood that weighs as much as a sphere 15 inches in diameter ?

PROPERTIES OF CUBES.

If a cube be divisible by 6, its root will also be divisible by 6. And if a cube, when divided by 6, has any remainder, its root divided by 6 will have the same remainder.

All exact cubes are divisible by 4, or can be made so by adding or subtracting 1.

All exact cubes are divisible by 7, or can be made so by adding or subtracting 1.

All exact cubes are divisible by 9, or can be made so by adding or subtracting 1.

Every cube is divisible by the cube of each of its prime factors.

ROOTS OF HIGHER POWERS.

When the exponent of a power can be resolved into two or more factors, by successively extracting the roots denoted by those factors, we may obtain the root desired. Thus, as 12=3×2× 2, the cube root of the square root of the square root of a number, is equal to the 12th root. So the square root of the square root the 4th root; the cube root of the cube root is the 9th root; the cube root of the square root is the 6th root, &c.

The demonstration of the following rule depends upon Algebraical principles, and therefore cannot properly be introduced here. In its application to square and cube roots, the student will be able to trace some analogy to the rules already given.

GENERAL RULE

FOR EXTRACTING THE ROOTS OF ALL POWERS.

At the left of the number whose root is required, arrange as many columns as are equal to the index of the root, writing 1 at the head of the first or left hand column, and zero at the head of each of the others.

Divide the number into periods of as many figures as the index of the root requires. Write the root of the left hand period as the first figure of the true root.

Multiply the number in the first column by the root figure, and add the product to the second column; add the product of this sum by the root figure to the third column, and so proceed, subtracting the product of the last column from the given number.

Repeat this process, stopping at the last column, and thus proceed, stopping one column sooner each time, until the last sum falls in the second column.

To determine the second root figure, consider the number in the last column as a trial divisor, and proceed with the second root figure thus obtained, precisely as with the first.

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