for abridging the work, Art. 41, and obtain the remaining figures of the root.

3. Extract the cube root of 9 to 9 decimals

4. What is the cube root of 15% to 5 decimal places?

Ans. 2.50222.

5. What is the cube root of 265 to 8 decimals?

Ans. 0.68278406.

6. What is the cube root of 0·0000031502374 to 13 decimals? Ans. 00146593403377. 7. What is the cube root of to 21 decimals? Ans. 0.793700525984099737376. 8. What is the cube root of to 10 decimals? Ans. 0.5227579585.

9. What is the cube root of to 7 decimals? Ans. 0.9032157.

10. What is the cube root of to 8 decimals?

Ans. 0.13992727.

EXAMPLES

INVOLVING THE

PRINCIPLES OF THE CUBE ROOT.

79. It is an established theorem of geometry, that all similar solids are to each other as the cubes of their like dimensions.

1. If a cannon ball 3 inches in diameter weigh 8 pounds, what will a ball of the same metal weigh, whose diameter is 4 inches?

By the above theorem, we have 3: 43: 8 pounds : 18 pound, for the answer.

2. Suppose the diameter of the sun to be 887681 miles; the diameter of the earth, 7912 miles. How many times greater in bulk is the sun than the earth?

(887681)=699472706450842241; (7912)=495289174528;

699472706450842241 ÷ 495289174528 = 1412251

times, nearly.

3. How many

of a cubic inch?

cubic quarter inches can be made out Ans. 64.

4 Required the dimensions of a rectangular box, which shall contain 20000 solid inches; the length, breadth, and depth being to each other as 4, 3, and 2.

2000 × × × 1 =250, whose cube root is 5

9'4103, nearly.

[9.4103×4=37-6412, length.

Ans. 94103 x 3=28.2309, breadth.

Or, as follows:

̧9·4103×2=18·8206, depth.

If we were to augment the width of this box, so as to make it as wide as it is long, its volume would become of 20000 26666. Again, if we augment the depth of this new box, so that it may be as deep as it is wide, and as it is long, its volume will become 2 times 266663 =533331, which is the contents of a cubical box, whose side is equal to the length of the original box. Hence, 53333=37641, nearly, for the length. The width is of this length, and the depth is this length.

3

5. What is the side of a cube which will contain as much as a chest 8 feet 3 inches long, 3 feet wide, and 2 feet 7 inches deep? Ans. 47-9843 inches.

6. Four ladies purchased a ball of exceeding fine thread, 3 inches in diameter. What portion of the diameter must each wind off so as to share of the thread equally?

Solution.

After the first one had wound off her share, the ball which remained would contain as much thread as it did in the first place. Therefore, its diameter was 362.72568 inches, nearly.

The diameter, after the second one had wound off her share, was 32=4=2·38110 inches, nearly.

The diameter, after the third one had wound off her share, was 3=2=1·88988 inches, nearly.

Hence, the portions of the diameter which they must wind off are as follows:

Inches, nearly.

The 1st lady must wind off 3·00000—2.72568=0·27432

ROOTS OF ALL POWERS.

80. WHENEVER the index denoting the root required is a composite number, the root can be found by successive extractions of the roots denoted by the prime factors of the original index.

Thus, the 4th root may be found by extracting the 2d root twice in succession.

The 6th root may be obtained by extracting the 3d root of the second root.

The 8th root may be found by extracting the 2d root three times in succession.

When the index denoting the root is a prime, we must have some direct method of obtaining the root.

By a similar train of reasoning, as was used in deducing the rule for the cube root, we determine, in general, for any root, the following

I. Point the number off into periods of as many figures each as there are units in the index denoting

the root.

II. Find, by trial, the figure of the first period, which will be the first figure of the root; place this figure to the left, in a column called the FIRST COLUMN. Then multiply it by itself, and place the product for the first term of the SECOND COLUMN. This, multiplied by the same figure, will give the first term of the THIRD COLumn. Thus continue until the number of columns is one less than the units in the index denoting the root.

Multiply the term in the LAST COLUMN by the same figure, and subtract the product from the first period, and to the remainder bring down the next period, and it will form the FIRST DIVIDEND.

Again, add this same figure to the term of the First COLUMN, multiply the sum by the same figure, and add the product to the term of the SECOND COLUMN; which, in turn, must be multiplied by the same figure, and added to the term of the THIRD COLUMN, and so on, till we reach the LAST COLUMN, the term of which will form the FIRST TRIAL DIVISOR.

Again, beginning with the FIRST COLUMN, repeat the above process until we reach the column next to the last. And so continue to do, until we obtain as many terms in the FIRST COLUMN as there are units in the index denoting the root; observing, in each successive operation, to terminate on the column of the next inferior order.

III. Seek how many times the FIRST TRIAL DIVISOR, when there are annexed to it as many ciphers, less one, as there are units in the index, is contained in the FIRST DIVIDEND; the quotient figure will be the second figure of the root. Then proceed with this figure the same as