i. e., the cube of every number composed of two parts is equal to the cube of the first part; plus the product obtained by multiplying the second part by the sum of three times the square of the first part, plus the sum of three times the first part plus the second multiplied by the second.

222. To extract the Cube Root.

What is the cube root of 259694072?

Solution.(a.) Since the number lies between 1,000,000 and 1,000,000,000, its root must lie between 100 and 1000, and hence must contain three figures. Separating it into periods of three figures each, it will take the form 259694072.

(b.) If we now let a represent the hundreds of the root and b the tens, it is evident that the whole of as will be found in the left hand period, and of (a + b)3 in the two left hand periods.

(c.) The greatest cube in 259 is 216, the root of which is 6. Therefore, 6 = = a, = the hundreds figure of the root. Subtracting a3, = 216 millions, from 259 millions leaves 43 millions, to which adding the next period gives 43694 thousands. We may regard this as a dividend, and it must contain [3 a2 + (3 a + b) × b] b, i. e., the remaining part of (a + b)3.

(d.) Now, as we know a, we can find 3 a2, and make use of it as a trial divisor to find b. But a being 6 hundreds, 3 a2 must equal 108 en-thousands, and as b is tens, 3 a b must be hundred-thousands. Hence, we may disregard the two right hand figures of the dividend, and see how many times the trial divisor, 108, is contained in 436.

(e.) The quotient being 4, we write it as the probable tens figure of the root, and have next to complete the true divisor, 3 a2 + (3a + b) b. But (3 a + b) = 18 hundreds + 4 tens = 184 tens, and (3 a + b) b 184 tens multiplied by 4 tens = 736 hundreds. Hence, the true divisor 3 a2 + (3 a + b) b : 108 ten-thousands + 736 hundreds

11536 hundreds.

=

(f) Multiplying this by 4 gives [3 a2 + (3 a + b) b] b = 46144 thousands, which, being greater than 43694 thousands, shows that there are not as many as 4 tens in the root, and that b is less than 4. (9.) Assuming b 3, and proceeding as before, we find that the 108 ten thousands +549 hundreds

=

=

=

true divisor 3 a2 + (3 a + b) b 11349 hundreds; and [3 a2 + (3 a + b) b] b : 34047 thousands, which, being less than 43694 thousands, shows that 3 is the true tens figure of the root. Subtracting 34047 thousands leaves 9647 thousands, to which adding the next period, 072 units, gives 9647072 for a new dividend, which must contain the remaining part of the power.

(h.) If we now let a' represent the part of the root already found,

i. e., the 63 tens, and b the units, we shall have 259694072 = a/s + [3 a/2 + (3 a' + b') b'] b', and, as we have already subtracted a, 9647072 will contain [3 a/2 + (3 a' + b') b'] b'.

(i.) But 3 a/2 3 times the square of 63 tens

11907 hundreds, which may, as before, be made use of as a trial divisor to find b'.

As 3 a2 is hundreds and b' is units, 3 a/2 b' must be hundreds; hence, no part of 3 a' b' can be found to the right of hundreds, and we may disregard the two right hand figures of the dividend, and see how many times the trial divisor, 11907, is contained in 96470.

(j) The quotient being 8, we write 8 as the probable units figure of the root, and complete the true divisor, 3 a2 + (3 a + b) b. 3 a + t 189 tens + 8 units 1898 units, and (3 a + b) b = 1898 X 8 = 15184. Therefore, 3 a2 + (3 a + b) b : = 11907 hundreds + 15184

units

=

1205884 the true divisor.

(k.) Multiplying this by 8 gives 9647072, which shows that 8 is the true value of b', and 638 is the root required.

Proof. See if 6383 =259694072.

(1.) Had the root contained another figure, we might have taken a'' to represent the part already found, and b'' to represent the next figur, when we should have, (a" + b'')3 = all3 + [3 aꞌꞌ2 + (3 a" + b'') b'] b" equal the number.

(m.) Much of the labor of finding the trial divisor, 3 a/2, might havo been avoided. For as a' = a + b, 3 a/2 must equal 3 times the square of a+b, or 3 times (a2+2 ab + b2) 3a26 a b + 3 b2.

=

(n.) But the previous trial divisor, 3 a2 + (3 a + b) b

3 ab+b2, and the number which stands above it equals (3 a + b) b = 3 ab+b2; hence the sum of these equals 3 a2+ 6 a b + 2 b2, which only lacks b2 of being equal to 3 a2 + 6 a b + 3 b2, or to 3 a". Hence, by squaring b, and adding it to this result, we have 3 a' = 3 a3 +6ab3 b2.

(o.) The work may be written thus:

a, b, b/ 259694072 (6 38

216000000

3 a2 = 1080000) 43694000

(3 a + b) b = 183 × 3 = 54900

630= a'

(p.) By keeping in mind the denominations, so as to render the zeros unnecessary, we should have the following form:

223. Rule for the Cube Root.

As a similar process will always apply, we may describe the method of extracting the cube root of a number thus: —

First. Divide the given number into periods of three figures each, beginning with the units.

Second. Find the greatest cube in the left hand period, and place its root as the first figure of the required root.

Third. Subtract this cube from the left hand period, and to the remainder bring down the next period, calling the result a dividend.

Fourth. Find three times the square of the part of the root already found, and make it a trial divisor.

Fifth. See how many times the trial divisor is contained in the dividend, excepting the two right hand figures, and write the quotient as the next figure of the root.

Sixth. To three times the part of the root previously found, annex the last root figure, multiply the result by the last figure, and placing the product under the trial divisor, two places to the right, add it to the trial divisor. This will give the true divisor.

Seventh. Multiply the true divisor by the last root figure, placing the product under the dividend.

Eighth. Subtract the product from the dividend, and to the remainder annex the next period for a new dividend.

Ninth. Add the square of the last quotient figure to the last true divisor and the number standing over it. The sum will equal three times the square of the root already found, and will be the second trial divisor. Tenth. Now proceed as directed from the fifth forward.

What is the cube root of each of the following numbers?—

224. Cube Root of Fractions.

(a.) The cube root of a fraction equals the cube root of its numerator, divided by the cube root of its denominator; and if both terms of the fraction are not perfect cubes, only its approximate root can be obtained.

(b) If in such cases the denominator is not a perfect cube, it will be well to multiply both terms by the square of the denominator, or by such other number as will make it so. Thus,

This differs from the true root by less than .

(c.) If a greater degree of accuracy is required, both terms may be multiplied by some perfect cube before extracting the root. Thus, –

which differs from the true root by less than 2.

What is the cube root of

225. Cube Root of Decimal Fractions.

(a.) In order that a decimal fraction may have a perfect cube for its denominator, it must contain 3, 6, 9, or some multiple of three places in

its numerator.

Thus, the denominators of .008, .027, .512, .003, and .375067 are each perfect cubes, while the denominators of .8, .08, .0027, and .56789 are not. (c.) If a decimal fraction, the root of which is required, does not contain 3, 6, 9, or some exact multiple of 3 decimal places, zeros must be annexed, so that the denominator may be in all cases a perfect cube.

1

100

=

.01, &c., it follows that there will be as many decimal places

in the fractional part of the root, as there are times three decimal places in the fractional part of the power. Hence, we carry out the root to as many decimal places as we choose, by annexing three zeros to the power for each additional figure we wish to obtain in the root.

What is the cube root to four places of decimals of

1. .37.

2. .6735.

3. 1.76.

4. 29.78.

5. .427.

6. .0007.

226. Rule for extracting a Root of any degree.

A root of any degree may be found as follows: First. Divide the given number into periods of as many figures each as there are units in the index of the required root.

Second. Find the greatest power of the required degree in the left hana period, and place its root as the first figure of the required root.

Third. Subtract the power from the left hand figure, and to the remainder bring down the first figure of the next period for a dividend.

Fourth. Raise the part of the root already found to a power one degree less than the given power, and multiply the result by the index of the required root, calling the result a trial divisor.

Fifth. Divide the dividend by the trial divisor, and the quotient will probably be the next figure of the root. To ascertain whether it is, place it in the root, and raise the number thus found to the required power. If the result equals the first two periods of the given number, or is less, the root figure is correct; but if, as will often be the case, it is greater, the root figure is too large.

Sixth. Having found the true root figure, find the remainder, and form a trial divisor, &c., as before. 29 *