slabs. But before multiplying, we will increase that sum by the surface of the second set of corner pieces, and of the second little cube. The length of each corner piece is the same as a side of the cube, figure 4, which is 720 feet; hence, adding 20 to 2140 already found, we obtain 2160, which, being 3 times 720, will be the linear extent of the three corner pieces. Were we to multiply 2160 by 6, we should find the surface of these three corner pieces, but as we wish also the area of one of the faces of the second little cube, we add 6 to 2160, and thus obtain 2166, which, multiplied by 6, will give 12996 for surface of second set of corner pieces and of second little cube; this added to 1555200, gives 1568196 for the surface of the whole second series of additions. Multiplying 1568196 by 6, we obtain 9409176 cubic feet, which have thus been added to the cube represented by figure 4; hence the cube whose side is 726 feet is the one sought. The above work may be arranged as follows: If we omi. the ciphers on the right, and omit unnecessary terms, the work will take the following condensed form: NOTE. In the extraction of the cube root, as just illustrated, it will be noticed that each divisor is a geometrical surface; that is to say, the product of two dimensions, width and breadth, for example; and of course the quotient must be the other dimension, that is, the thickness. But it is important to remember that it is only squares and cubes, square roots and cube roots, that can have any relation to geometrical dimensions; any higher power of a number as 45, or any other root as, cannot be illustrated by blocks. The principle, therefore, of involution and evolution is, strictly speaking, independent of surfaces and solids; it is purely arithmetical. From the foregoing demonstration we may deduce the following RULE. I. Separate the number whose root is to be found, into periods of three figures each, counting from the units' place towards the left. When the number of figures is not divisible by 3, the left-hand period will contain less than 3 figures. II. Seek the greatest figure whose cube shall not exceed the first or left-hand period; write it after the manner of a quotient in division for the first figure of the root. Place this figure for the head of a first left-hand column, and its square for the head of a second left-hand column, and subtract its cube from the first period. To the remainder bring down a second period for the FIRST DIVIDEND. Add the figure in the root to the term of the IST COLUMN already found, for its next term, which multiply by the same figure, and add the product to the term already found in the 2D COLUMN, for its next term, which will be a TRIAL DIVISOR. III. Find how many times the trial divisor, with two ciphers annexed, is contained in the dividend; write the quotient for the next figure of the root. Annex this figure to the last term of the 1ST COLUMN, after having added to that term the preceding quotient figure; this will give the next term of the 1ST COLUMN. Multiply this term by the last found figure in the root, and add the product, after advancing it two places to the right, to the last term of the 2D COLUMN, for its next term. Multiply this term by the last found figure of the root, and subtract the product from the dividend, and to the remainder bring down the next period for a NEW DIVIDEND. Proceed as before until all the periods have been brought down. NOTE 1.-When any dividend is not so great as the corresponding trial divisor with two ciphers annexed, write 0 for the next figure of the root, and to the dividend bring down the next period. Use the same trial divisor as before, but with four ciphers annexed. NOTE 2.-The trial divisor, being less than the true divisor, will sometimes give too large a quotient figure; when the multiplication of the true divisor by this figure shows such to be the case, this figure must be made smaller. NOTE 3.-By the above rule, which is different from the rule usually given by the aid of geometrical diagrams, we have managed to keep distinct all the geometrical magnitudes; thus our first column represents the numerical values of lines, the second column represents the numerical values of surfaces, and the third column corresponds to solids. And, as we are never required to multiply by any number greater than is expressed by a single digit, the labor of multiplying and adding results to the terms of the successive columns is far simpler than at first might be supposed. By means of these auxiliary columns the work bears a close analogy to Horner's method of solving numerical cubic equations. (See Treatise on Algebra.) The use of auxiliary columns be comes very apparent in the extraction of roots of the higher orders, as the fifth root, the seventh root, &c. 2. What is the cube root of 10077696? 4. What is the cube root of 42875 ? CASE II Ans. 216. Ans. 1331. Ans. 35. Ans. 49. Ans. 1936. To extract the cube root of a decimal fraction, or of a number consisting partly of a whole number and partly of a decimal, we have this RULE. I. Annex ciphers to the decimals, if necessary, so that they may be separated into equal periods. II. Separate the decimals into periods of 3 figures each, counting from the decimal point toward the right, and proceed as in whole numbers. NOTE.-If the given number has not an exact root, there will be a remainder after all the periods have been brought down. The process may be continued by annexing ciphers for new periods. EXAMPLES. 1. What is the cube root of 0.469640998917? Ans. 0.7773. 2. What is the cube root of 18-609625? Ans. 2.65. 3. What is the cube root of 1.25992105? 4. What is the cube root of 2? 5. What is the cube root of 9? 6. What is the cube root of 3? Ans. 1.08005. Ans. 1.2599. Ans. 2.08008. Ans. 1.4422. CASE III. To extract the cube root of a vulgar fraction, or mixed number, we have this RULE. I. Reduce the fraction or mixed number, to its simplest fractional form. 11. Extract the cube root of the numerator and denomina tor separately, if they have exact roots, but when they have not, reduce the fraction to a decimal, and then extract the root by Case II. |
