side of a stream which flows at the foot of the hill is 400 feet: what is the width of the stream? 402. Table, showing the third power or cubes of units, tena and hundreds. By the above it will be seen that the cube root of any whole number, composed of one, two, or three figures, must be units; of four, five, or six figures, must be units and tens; of seven, eight, or nine figures, must be units, tens and hundreds; and hence, generally, that if we point a number off into periods of three figures each, beginning with the units, the number of figures in the cube root will be indicated by the number of the periods. The left hand period may contain but one or two figures. NOTE I.- The principle above elucidated applies to decimal fractions; but every period in decimal fractions must contain three figures. 403. Before extracting the cube root, let us involve 64 to the third power, and preserve the separate products. We have already seen (Art. 388) that the square or second power of 64 is 602 + 2 × (60 × 4) + 42. By multiplying this square by 64 (60+ 4), we shall obtain the third power of 64. 642602+2 × (60 × 4) + 43 64 = 60 + 4 603 +2 × (60 × 60 × 4) + (60 × 42) In this product we have four distinct parts, as follows: (1), 603, the tens raised to the 3d power, =216000 (2), 3 × (602 × 4), 3 × the square of the tens X the units, 43200 (3), 3 × (60 × 42), 3 X the tens X the square of the units, = (4), 43, the units raised to the 3d power, 2880 = 64 262144 Thus we see that a cube whose root is composed of tens and units, contains (1) the cube of the tens, (2) three times the square of the tens multiplied by the units, (3) three times the tens multiplied by the square of the units, and (4) the cube of the units. 404. Observing, now, that the first of these parts is the cube of the tens, and that the units is a factor in each of the other parts, we will proceed to extract the cube root in the following ILL. EX. What is the cubę root of 264609.288? 3x sq. of tens (trial divisor), 3x tens X units, Square of units, True divisor, SOLUTION. 11536 × 4 46144 Pointing the number off into periods of three figures Bach, by placing a dot over the units and every third figure to the right and left, we find that the root will consist of three figures, and the cube of the tens must be contained in 264(000). The largest cube contained in 264(000) is 216(000), the root of which is 6(0); this we write the tens' term of the root, and subtract its cube, 216(000), from 264(), and to the remainder, 48(000), bring down the next period, 609, We know that this remainder, 48609, contains three times the square of the tens (the term already found), multiplied by the units; and though it contains other terms, since this is much the largest, we take 3 times 602 (three times the square of the tens) for a trial divisor, and, dividing 48609 by it, obtain 4 for the units' figure. We multiply 3 × the tens (60) by the new term of the root (4), and place the product under the trial divisor, and under this, place the square of the units' figure; and thus form our true divisor, consisting of the last three parts of a perfect cube (Art. 403), wanting the units as a factor in each. Multiplying their sum by 4, we have 46144, This we subtract, and to the remainder bring down the next period. Considering 64 as the tens in the root, we multiply its square (6402) by 3 for a new trial divisor, and, proceeding as before, obtain 64.2 for the cube root of 264609.288. From the above we deduce the following 405. RULE For extracting THE CUBE ROOT. Point off the given number into periods of three figures each, by placing a dot over the units, and every third figure to the left in whole numbers, and to the right in decimals: Find the greatest cube in the left hand period, and write its root as the first term in the answer. Subtract the cube from the left hand period, and to the remainder bring down the next period for a dividend. Multiply the square of the root already found, considered as tens, by three for a trial divisor. Divide the dividend by the trial divisor, and place the result as the next term in the root. To the trial divisor add three times the former term in the root (considered as tens), multiplied by the last term, also the square of the last. Multiply this sum by the last term, and subtract the product from the dividend. To the remainder bring down the next period for a new dividend. Multiply the square of the terms of the root already found (consid ered as tens), by three for a trial divisor, with which divide and proceed as before. NOTE I. - When a zero occurs in the root, annex two ciphers to the trial divisor, and, bringing down another period, proceed as before. NOTE II. - If a root figure proves too large, substitute a lower, and repeat the work. NOTE III. -When a remainder occurs after all the periods are brought down, the root may be more nearly approximated by annexing periods of zeros, and continuing the operation. NOTE IV. The cube root of a common fraction may be obtained by extracting the root of each of the terms when they are perfect cubes; when they are not, the fraction may be reduced to a decimal. NOTE V.-Mixed numbers may be reduced to decimal fractions, or to improper fractions when the denominator of the fractional part is a cube number. 406. The above rule may be illustrated by means of blocks. A cube number represents a cube, the edge of which is the cube root of the number. Let there be a cube of 262144 solid inches, whose edge we wish to determine. Having found by pointing and trial that the greatest cube of tens in 262144 is 216(000), the root of which is 6 tens, we will let 216000 inches be represented by the following diagram (Fig. 1), having for its edge 6 tens of inches, or 60 inches. one side of this cube will be 6023600 inches, and upon three sides 10800 inches. Using this as a trial divisor, we find the thickness of the ad litions to be 4 inches. The additions ar represented by Fig. 2 These additions being made, the soli w be represented by Fig. 3 64 in. To complete the cube, it also requires three oblong rectangular blocks, whose length is 60 inches, and whose end is 4 inches square (Fig. 4); also a cube, whose edge is 4 inches (Fig. 6). The side of one of the oblong blocks being 60 x 4, one side of the three will be 3 times 60 x 4720 square inches, and one side of the small cube will be 42 16 square inches. If, now, we multiply the sum of these surfaces, 10800+720+16, 11536 (Fig. 7), by their thickness, 4, and increase the cube 216000 by the product, we form a perfect cube (Fig. 8), whose edge is 64 inches. And since there is no remainder, 262144 is a perfect cube, of which 64 is the root. 4 in. |