Cube Root, to which work we shall, in the following, make reference. For the sake of simplicity we will suppose we are required to find the number of feet in the side of a cube, so that its volume shall be 387420489 cubic feet.

We know that the number denoting the cube root of 387420489, or the number of feet in the side of our sought cube, must consist of three places of figures, since the number 387420489 is greater than the cube of 100, and less than the cube of 1000. The greatest cube which does not exceed the left-hand period, 387, is 343, hence, the figure in the hundredth's place is 7, that is, the side of the cube sought is greater than 700, and less than 800 feet. If we take the cube whose side is 700 feet, the area of one of its faces will be (700)2 = 490000 square feet, and its volume will be 490000×700= 343000000 cubic feet, which taken from 387420489 leaves 44420489 cubic feet. Hence, it is necessary to increase the side of this cube, so that its volume may be augmented by 44420489 cubic feet, which is called the first dividend. Now we know that a cube may be increased by three equal rectangular parallelopipedons, which we have called slabs, (see page 315, of this Appendix,) three other equal rectangular parallelopipedons called corner pieces, and a little cube; and when thus increased it will still be a perfect cube. Since the volume of the three slabs is by far the largest portion of the additions, and since the surface of each slab is the same as the face of the cube, viz. (700) =490000 square feet, the three together will be 3× (700)2 =1470000 square feet, which has been called the first trial divisor, it therefore follows, that if we divide 44420489 by 1470000, we shall obtain the approximate thickness of these slabs. Performing this division, we find a quotient between 30 and 40, but as the divisor so used is always less than the surface of all the additional parts, by the surface of the three corner pieces and little cube, our quotient may sometimes be too large, it can never be too small; in this case, 30 is too large, therefore we must consider the true quotient as comprised between 20 and 30, so that the figure in the ten's place is 2.

Hence, the thickness of the slabs is 20 feet, this is also the width of the corner pieces, their lengths being the same as the side of the cube, 700, so that the surface of the three corner pieces is 3 × 700×20=42000. Also the side of the little cube is 20, one of its faces is (20)2 = 400, hence, the entire surface of all the additions is 1470000+42000+400=1512400 equal the true divisor, which being multiplied by 20 gives 30248000 cubic feet, which subtracted from 44420489, leaves 14172489 cubic feet, which still remain to be added to the cube whose side is 720.

Using, as before, three times the square of 720, which is 3x (720)2 = 1555200, which is the surface of the three new slabs, making a second trial divisor, which we find is contained in 14172489 between 9 and 10 times, therefore the unit figure of our root is 9, which is the thickness of the new slabs, and the width of the new corner pieces, also the side of the new little cube, hence, the entire surface of the additions is 3 × (720)2 +3×(720)× 9+92=1574721 equal true divisor, which multiplied by 9 gives 14172489 for the whole number of cubic feet added this second time, which being subtracted from the second dividend, 14172489, leaves no remainder, so that

a cube whose side is 729 feet, contains just 387420489 cubic feet, and is therefore the side of the cube sought

Had the number, whose root it was required to find, been such as to give more than three places of figures in the root, the process would not have differed from the above, only being more lengthy, each additional figure being found by means of a trial divisor which is three times the surface of a face of the cube already obtained. Each successive operation, when a new figure is found, augments the cube last obtained by three slabs, three corner pieces, and a little cube. In the arrangement of the work under Example 1, which has already been referred to, we have omitted the zeros on the right, for the sake of simplicity; for the same reason, the work is arranged in two distinct columns, by which means the operation of multiplying, &c., is made quite simple. By means of these auxilliary columns the work bears a close analogy to the new method of solving numerical cubic equations, as given by Mr. Horner, of Bath, England. The use of auxiliary columns becomes very apparent in the extraction of roots of the higher orders, as the fifth root, the seventh root, &c. The above rule for the cube root may be beautifully illustrated by means of the ordinary blocks prepared for the common rule.

In squaring a number, we multiply the right-hand digit into itself, thus obtaining the right-hand digit of the product, or of the square; hence, the right-hand digit of a perfect square must arise from squaring some one of the digits. The right-hand digit of the squares of the nine digits will end with one of the following digits, 1, 4, 5, 6, or 9. Therefore every square number must end with one of the digits 1, 4, 5, 6, 9, or else with an even number of zeros. Hence, any number terminating on the right with either of the digits 2, 3, 7, 8, or with an odd number of zeros, cannot have its square root accurately found.

The right-hand digit of the cubes of the nine digits will end with all the different digits, so that we cannot, by the terminal figure of a number, pronounce that its cube root cannot be found, except when the number ends with a number of zeros which is not a multiple of 3, in which single case we know that the cube root cannot be accurately found.

When a number is divided by 5, we shall obtain, for a remainder, one of the following numbers, 1, 2, 3, 4, unless the number is a multiple of 5, in which case it is exactly divisible by 5.

If a number is divisible by 5, its square will obviously be divisible by 5.

If a number give 1 for a remainder, when divided by 5, it may be considered as composed of a multiple of 5, plus 1; consequently when squared by the method explained under this article, page 311, its first and second terms will be multiples of 5, and the third term will be the square of 1, so that when a number is composed of a multiple of 5 plus 1, its square will give the same remainder, when divided by 5, as will be found by dividing the square of 1 by 5, which remainder is therefore 1.

And for a similar reason, when a number is composed of a multiple of 5, plus 2, the remainder found by dividing its square by 5, is the same as the remainder found by dividing the square of 2 by 5, which remainder is 4.

If a number is composed of a multiple of 5, plus 3, the remainder found by dividing its square by 5, will be the same as the remainder found by dividing the square of 3 by 5, which remainder is 4.

If a number is composed of a multiple of 5, plus 4, the remainder found by dividing its square by 5, will be the same as the remainder found by dividing the square of 4 by 5, which remainder is 1.

From which we discover, that any square number being divided by 5, will give either 1 or 4 for a remainder, unless it is a multiple of 5.

Hence, every square number is either a multiple of 5, or else a multiple of 5 increased or decreased by 1.

Any number divided by 7, will give one of the following remainders, 1, 2, 3, 4, 5, 6, unless it is a multiple of 7, in which case it is obviously divisible by 7.

When a number is not a multiple of 7, it may be regarded as composed of two parts, the first of which is a multiple of 7, plus one of the remainders, 1, 2, 3, 4, 5, 6.

Now, when a number of this kind is cubed by the method explained on page 315 of this Appendix, its cube will consist of four parts, namely, the cube of the first, plus three times the square of the first into the second, plus three times the first into the square of the second, plus the cube of the second; the first three of these parts contain a multiple of 7 as a factor, hence, the remainder will be found by dividing by 7 the last part, which is the cube of one of the numbers 1, 2, 3, 4, 5, 6. The cubes of 1, 2, 3, 4, 5, and 6, are respectively 1, 8, 27, 64, 125, and 216, dividing these cubes by 7, we find either 1 or 6 for the remainder.

From which we see that any cube number being divided by 7, will give either 1 or 6 for a remainder, unless it is a multiple of 7.

Therefore, every cube number is either a multiple of 7, or else a multiple of 7 increased or diminished by 1.