period for a dividend. Find a divisor by multiplying the square of the root by 300, see how often it is contained in the dividend, and the answer gives the next figure in the root; multiply all the figures in the root by 30, except the last, and that product by the square of the last. Cube the last figure in the root; add these three last found numbers, and subtract this sum from the dividend; to the remainder bring down the next period for a new dividend, and proceed as before.

The following rule taken from Dr.Hutton's Arithmetic, page 252, will be preferable, when the root consists of more than 3 figures.

Divide the given number into periods of 3 figures each, beginning at the unit's place, and take the root of the greatest eube in the left-hand period, for the first figure of the root, subtract its cube from the first period, and to the remainder bring down another period for a resolvend.

Multiply the square of the figure found by 3 for a divisor, by which divide the resolvend, neglecting the two right-hand figures, the quotient is to be tried for another figure of the root. To three times the former part of the root, annex this trial figure, and multiply the sum by it, and place the product

below the divisor, but two figures farther to the right, add those two lines together to complete the divisor; then multiply it by the trial figure, and subtract the product from the resolvend, and to the remainder bring down another period for a new resolvend.

Below the complete divisor, place the square of the figure of the root last found, and add it and the two lines above it, and the sum is a new divisor, with which proceed as before.

Find the cube or third roots of the following numbers respectively

If the given number is not a perfect cube,* the root may be found to any degree of exactness, by adding periods each consisting of 3 ciphers, but as this is an operation which is extremely tedious and troublesome, the following rule is preferable in such cases.

RULF. By trials, take the nearest cube to the given number either greater or less and call it the assumed cube. Then say by the rule of Proportion, as the sum of the given number and double the assumed cube, is to the sum of the assumed cube and double the given number, so is the root of the assumed cube, to the root required nearly.

Again, by using in like manner the cube of the root last found as a new assumed cube, another root will be found still nearer, and so on as far as we please, always using the cube of the root last found for the assumed cube.

Examples.

What is the cube root of 321 ?

Here the root found by trial is nearly 6,8
6,8314,432

Then as 314,432 ×2+321; 321×2+314,432:: 6,8:6,84702
In general it will be best to find the three first figures of the
root by the common rule, and then apply this rule to the result.
Find the cube or third roots of the following numbers?

To extract the cube root of a vulgar fraction, reduce the fraction to its lowest terms, then extract the cube root of the numerator for a new numerator, and the cube root of the denominator for a new denominator; but if these terms will not extract even, multiply the numerator by the square of the denominator, and the cube root of the product divided by the denominator, will give the root required. Or reduce the fraction to a decimal, and then extract the root.

* John Walker, in his Philosophy of Arithmetic, page 197, gives directions for easily finding the Cube Root, any Perfect Cube up to one billion, to which and to Dr. Hutton's Mathematics, vol. I, pages 85 to 88, I refer the reader.

Examples.

Extract the cube root 2 29 ?

The cube root of 729 is 9, and of 1728 is 12, therefore

is the root required.

Extract the cube root of ?

Here 2×25 50 and

root required.

50=3,6840315+5=,7368063=

Find the cube or third roots of the following fractions and mixed numbers respectively?

This

is the sign of the Cube Root, and signifies that the cube roof

of the number before which it is placed, is to be extracted.

QUESTIONS

FOR EXERCISE IN THE FOREGOING RULES.

A's money

1. A and B traded together, and gained 3601. multiplied by 3, was equal to B's multiplied by 2; what was the share of each?

2. Three merchants, A, B, and C, have gained 2347, which they are to divide so, that when the profit of A is multiplied and that of C by 4, their products are each of them to receive?

by 2, that of B by 3, equal; how much is

3. Suppose I owe in Birmingham for hardware, per invoice 2541 12s 6d the rider having called on me for payment, which would it be better for me to pay him in guineas, which I must purchase at 74d. each, or give him a bill on London, at 11 per cent. exchange, and what is the difference on this transaction?

4. Two men bought a quarter share in a lottery ticket, which turned out a prize of 20,000 for the whole ticket: they agreed to divide equally with a third person, not concerned therein; now every 1007 British being worth 108/ 6s. Ed. in Ireland, what is each man's share, allowing the lottery agent 5 per cent?

5. Shipped for Bristol a cargo of wheat, which sold there for 15601. The charges came to 107 10s. and my factor. charges me 4 per cent. commission; what will it amount to in Irish currency, exchange being 9 per cent?

6. A piece of ground was let to two partners, at a rent of 21 15s. per acre; but one part of it was better than the other. They divide it according to the quality of the ground, one paying 3 15s. for his part, and the other but 17 15s. per acre for his. Some time afterwards the landlord agrees to make an abatement of 11 per acre; what per acre ought each man now to pay ?

7. Two men go in partnership with a stock of 3001 each ; one to have of the profits, and the other. At the end of 3 months, he that had but of the profits advanced 2001 more; they gain in the course of the year 3751; how is this sum to be divided between them?

8. Four persons hire a coach to go 120 miles, for 81, with condition if they take up any person on the road, the profit