Required the square root of the following fractions, so near, that the square of each root may differ from the fraction which gave it by less than a millionth.

11.

12. V

,9128709
,8164965

14. VII 15.

8 =,8528028

=,2773500

13.,8819171 16. 1,0316385

17. Having ten square lots of ground, the first containing one acre, the second, two, the third, three, &c., to the tenth inclusive; required the length, in ft. and inches, of one side of each.

Ans. No. 1. 208 ft. 8 in. +

No. 6.

511 ft. 3 in.

No. 2.

295 ft. 2

in.

No. 7.

552 ft. 2 in. +

590 ft. 4 in.

626 ft. 11 in +

No. 5. 466 ft. 8

in. +

No. 10.

660 ft.

18. Having 6 acres of ground which we wish to plant with apple-trees, in equidistant rows, at right angles to each other, at what distance must the trees be placed, so that we may have 20 on the first acre, 30 on the second, 40 on the third, 50 on the fourth, 60 on the fifth, and 70 on the sixth? Ans. On the 1st acre at 46 ft. 8 in. +

19. Wishing to plant 529 trees in equidistant rows, at right angles to each other, so that the orchard may form a square, and have 35 trees to the acre; we would know the number of rows, the number of trees in each row, the distance, in feet and inches, from tree to tree, and how much land the orchard will contain. Ans. 23 rows; 23 trees in each; 35 ft. 3 in. + apart, and 15 A. 18 P. + in the orchard.

EXTRACTION OF THE

SECTION XX.

CUBE ROOT-EXTRACTION OF THE

ROOTS OF ALL POWERS.

Extraction of the Cube Root.

375. The cube of a number being a product in which the number is three times factor, the cube root of a number is the number, which being written three times and involved, produces the given number.

Now the cube of one figure can never consist of more than three figures, (121,) neither can the cube root of a number expressed by three figures consist of more than one figure; for, if we suppose it to consist of two, its square (121) must contain at least three, and its cube four or five. The cube root, therefore, of a number, which is a perfect cube, and which does not contain more than three figures, must be one of the nine digits, and may easily be found by the table of powers. (357.)

376. If the cube consists of more than three figures, the root will contain two or more, because the product of two numbers cannot contain more figures than both factors. Now, if the root consists of two or more figures, we may consider it as a binomial, that is, as consisting of tens and units; then, calling the tens a, and the units b, we shall cube a + b by multiplying its square, which is already known, by a + b, thus :

a2+2ab+b2

a + b

a3+2ab+ ab3

a2b+2ab2 + b3

a3 +3a2b+3ab2 + b3

from which we find, that the cube of a number, consisting of tens and units, contains the cube of the tens, three times the square of the tens multiplied by the units, three times the tens multiplied by the square of the units, and the cube of the units. By this formula, which shows the constituent parts of the cube, the cube root is extracted.

377. We may also verify this formula arithmetically, by applying its several parts to the dimensions of a foot; namely, (102) inches, thus:

The sum of the several parts being 1728 cubic inches, the true content of a cubic foot, shows the correctness of the formula.

378. We may also verify the formula geometrically by constructing a cube according to its several parts, thus:

Let a be represented by the following cube:

3ab, by the three following parallelepipeds:

3ab, by the three following parallepipeds:

and 3 cubed by the following cube:

All these solid figures combined, form the following figure, which is the cube of a + b:

To extract the cube root of a number by means of the formula just established, we reason as follows: Every number, consisting of two or more significant figures, may be considered binomial, and the cube of its tens-that is, the cube of the number on the left of its unit figure-must be a number of thousands; because 10s 1000, and therefore cannot form any part of the three right-hand figures of the cube. Again, if the part of the root, considered as tens, contains more than one figure, the part on the left of its right-hand figure is a number of tens with regard to that figure, or a number of hundreds, with regard to the first; therefore, as 100=1000000, the cube of this part, being a number of millions, cannot form any part of the first six figures of the cube. Continuing to reason thus, it is evident, that in order to find that part of the cube which contains the cube of the left-hand figure of the root, we must point off the figures of the cube into periods of three figures each, beginning at the right-hand. This, therefore, is the first step.

Again, the addition of the minor parts of the cube to the cube of the tens can never increase the root of the tens by a unit, because this unit would be 10, and the difference between a3 and (a+10)3 is 3a2 × 10 + 3a × 10o + 103, which, as 10 is greater than b, is greater than 3a2b+3ab + b3. Therefore, having pointed off the given number, to find the first figure, or tens of the root, we extract the root of the greatest cube contained in the left-hand period of the given number, and place it as we do a quotient figure in division. This is the second step.

We cube the root, and subtract its cube from the left-hand period of the given number, and, to the right of the remainder, we bring down the next period of the cube. The number thus formed is called the new dividend. This is the third step.

Having subtracted the cube of the tens, or a3, the remaining part of the cube contains 3a2b+3ab3 +b3. We must therefore endeavour to find these three parts, add them together, and subtract their sum from the new dividend. Now the root a is a number of tens, and as 3ab is a number of hundreds, it cannot form part of the two right-hand figures of the new dividend. We therefore point off those two figures, and consider the part on the left as containing 3ab; that is, the triple square of a multiplied by b. Then, as the product of two numbers, divided by one of them, gives the other, we square the root, multiply its square by 3, and place the product for a divisor, on the left of the new dividend. We then divide the left-hand

part of the new dividend by the divisor, and place the quotient or unit figure b, in the root, on the right of a. This is the fourth step.

We next draw a line under the new dividend, and, underneath this, we place the remaining parts of the cube in order, as we find them from the formula. We first multiply the divisor by b, and, because b is a number of hundreds, we place two ciphers on the right, which evidently gives 3ab, and place the product under the line. We then square b, multiply the square by 3a, and placing a cipher on the right, because a is a number of tens, we write this product under the other. Lastly, we cube b, and having written its cube under the line, we add the three parts together, and subtract their sum from the new dividend. As this completes the cube of the two figures of the root already found, it may be called the last step.

If the sum of the three parts should exceed the new dividend, b is too great, and must be diminished till the subtraction becomes possible. Also, the subtraction shows whether the given number is, or is not, the exact cube of the root obtained. For, as the sum of the minor parts completes the cube of the root, if this agrees with the new dividend, that is, if the subtraction leaves no remainder, the given number is the exact cube of the root found. If the subtraction leaves a remainder, the given number is not the exact cube.

379. We will now cube 27, and then, by the formula, extract its root. Thus :

27 × 27 × 27=9X9X9X9× 3 = 6561 × 3 = 19683

This operation, having been sufficietly explained in the preceding article, needs no farther development. There is, however, a difficulty, too obvious to be neglected, which is, that as the divisor, which is 12 hundreds, is contained 9 times in the 116 hundreds of the new dividend, the scholar would, if not farther assisted, naturally place 9 in the quotient, and would,