(2) 72 sq. ft.

(4) 3245704 sq. yds.
(6) 8.158 ft.

(8) 3000 sq. ft.

(10) £78 75. 0'636d.

(12) 7854 : 7122.
(14) 681 links.

(16) 5 inches.
(19) 4 ft.

(21) 66990 sq. ft.
(24) 560 sq. ft.

(26) 491′934 + sq. yds.
(29) £3 os. 10°422d.

(37) 4'472136 ft., or 25 ft. (38) Breadth, 9 in.; length, 12 in.

(48) 1783 poles; 70°624 + poles. (49) 9 ft. 8 in.

ARITHMETIC.

(MALES.)

EVOLUTION. CUBE ROOT.

To Extract the Cube Root of a Given Number. The rule for extracting the cube root of a given number is deduced from the method employed to extract the cube root of an algebraical expression. (See Algebra, Cube Root.)

Rule. Divide the number into periods of three figures each by placing a dot over the units' figure and over every third figure to the left in integers, and to the right in decimals. Each period is supposed to end in a figure having a dot over it. One or two ciphers may be added to the decimal figures, if required to complete a period.

Now take the first period to the left, which may consist of one, two, or three figures, according as it has a dot over the first, second, or third figure. Find the greatest cube number in that first period, place its cube root in the quotient on the right is in long division, and set the cube number itself under the first eriod. Subtract the cube number from the first period, and to .he remainder annex the second period for the next dividend.

The figure in the quotient is the first figure of the root required. Multiply its square by 3 and set down the product to the left of the dividend just obtained, as a trial divisor. Find how many times this divisor is contained in the number obtained by omitting the two last figures to the right in the dividend; place the result in the quotient for the second figure of the cube root. Annex two ciphers to the divisor. Under the divisor thus increased set down the number obtained by multiplying the first figure of the root by 3, annexing the second figure of the root, and multiplying the result by the second figure of the root. Add together the two numbers thus placed under each

other, and the sum will be the complete divisor. Multiply the latter by the second figure of the quotient, place the product below the dividend, and subtract it therefrom. (If the product be greater than the dividend, the former must be diminished by taking a lower number as the second figure of the root.)

To the remainder bring down the figures of the third period to form a new dividend. Then multiply the square of the quotient by 3, set down the product to the left of the dividend as a trial divisor, and proceed as before.

Three times the square of the quotient may be obtained by adding together; (a) the former complete divisor, (b) the number obtained by multiplying the first figure of the root by 3, annexing the second figure of the root, and multiplying the result by the second figure of the root, and (c) the square of the second figure.

Continue the process as long as any periods remain to be brought down.

If there should be a remainder after all the periods in the original number have been dealt with, the operation may be continued decimally by annexing three ciphers for each additional period till a sufficiently accurate result has been arrived at. (1) Find the cube root of 76009496256.

Here we place a dot over the units' figure, 6, and over the third figures to the left, 6, 9, 6, thus dividing the number into the four periods 76; 009; 496; 256.

The first period is 76. The greatest cube number it contains is 64, which is the cube of four. We place the root, 4, in the quotient, and the cube number 64 under the first period. Subtracting, we obtain the remainder 12, to which we annex the second period, 009, thus obtaining 12009 for a dividend.

In the quotient we have 4, which is the first figure of the root. The square of 4 is 16; multiplying 16 by 3 we get 48, which is placed to the left of 12009 as a trial divisor. Omitting the last two figures of 12009 we have 120. In 120 the trial divisor 48 is contained twice, we therefore place 2 in the quotient as the second figure of the root. Annexing two ciphers to 48 we get 4800. Then if we multiply 4, the first figure of the root, by 3, and then annex 2, the second figure of the root, to the product, we obtain 122; this multiplied by 2, the second figure of the root, becomes 244. Adding 244 to 4800 we get 5044, which is the complete divisor. The product of 5044 multiplied by 2 is 10088, which, subtracted from 12009, leaves the remainder 1921.

To this remainder we bring down the next period, 496, and thus obtain the next dividend, 1921496.

The new trial divisor
This is found most

The number in the quotient is now 42. will be 3 times the square of 42, or 5292. readily thus add together the last complete divisor, 5044, the number above it, 244, and the square of the second figure of the root, 22, or 4. Thus 5044 +244 + 4 = 5292. This number is contained 3 times in 19214; hence 3 is the third figure in the

root.

We then proceed as before. Annexing two ciphers to 5292 we have 529200. Then 42 × 3 = 126. Annexing the third figure, 3, we have 1263; 1263 × 3 = 3789. Adding, we have the complete divisor 532989.

The next trial divisor is found in a similar manner to the preceding ones, thus: 532989 +3789 +32=536787.