The manner of arriving at these results is to represent the unit by a letter, and expand the simple literal terms, and afterwards substitute their values in the result. (Art. 68.) If we expand (a-b) in place of (a+b,) the exponents and coefficients will be precisely the same, but the principles of multiplication of quantities affected by different signs will give the minus sign to the second and to every alternate term. Thus the 6th power of (a-b) is a®—6a3b-+-15aab2—20a3b3+15a2ba—6ab3+bo. (Art. 69.) This method of readily expanding the powers of a binomial quantity is one application of the "binomial theorem," and it was thus by induction and by observations on the result. of particular cases that the theorem was established. Its rigid demonstration is somewhat difficult, but its application is simple and useful. Its most general form may arise from expanding (a+b)". When n=4, we can expand it; When n= any whole positive number, we can expand it. Now let us operate with n just as we would with a known number, and we shall have We know not where the series would terminate until we know the value of n. We are convinced of the truth of the result when n represents any positive whole number; but let n be negative or fractional, and we are not so sure of the result. To extend it to such cases requires deeper investigation and rigid demonstration, which it would not be proper to go into at this time. We shall therefore content ourselves by some of its more simple applications. EXAMPLES. 1. Required the third power of 3x+2y. We cannot well expand this by the binomial theorem, because the terms are not simple literal quantities. But we can assume 3x=a and 2y=b. Then 3x+2y=a+b and (a+b)3=a3+3a2b+3ab2+b3 Now to return to the values of a and b, we have, Hence (3x+2y)3=27x3+54x2y+36xy2+8y3. 2. Required the 4th power of 2a2—3. Let x=2a2 y=3. Then expand (x-y)', and return the values of x and y, and we shall find the result, 16a-96a6+216a-216a2+81. 3. Required the cube of (a+b+c+d). As we can operate in this summary manner only on binomial quantities, we represent a+b by x, or assume x=a+b, and y=c+d. Then (x+y)3=x3+3x2y+3xy2+y3. Returning the values of x and y, we have (a+b)3+3(a+b)2(c+d)+3(a+b)(c+d)2+(c+d)3. Now we can expand by the binomial, these quantities contained in parenthesis. 4. Required the 4th power of 2a+3x. Ans. 16a+96a3x+216a2x2+216ax2+81x1. 5. Expand (+3y2)5. Ans. x1o+15x3y2+90x®y1+270x1y®+405x2y+243y1o. 6. Expand (2a+ax)3 Ans. 8a6+12æ3x+6a2x2+a3x3 Ans. x-6x+15x-20x+15x2-6x+1. 7. Expand (x-1).. EVOLUTION. CHAPTER II. (Art. 70). Evolution is the converse of involution, or the ex traction of roots, and the main principle is to observe how powers are formed, to be able to trace the operations back. Thus, to square a, we double its exponent, (Art. 65), and make it a2. Square this and we have a1. Cube a2 and we have ao. Take the 4th power of x and we have x1. The nth power of 3 is Now, if multiplying exponents raises simple literal quantities to powers, dividing exponents must extract roots. Thus, the 2 square root of a1 is a2. The cube root of a2 must be a3. The cube root of a must have its exponent, (1 understood,) divided by 3, which will make a3· Therefore roots are properly expressed by fractional expo nents. 5 The square root of a is a2, and the exponents, 3, 1, }, &c. indicate the third, fourth, and fifth roots. The 6th root of x is x; hence we perceive that the numerators of the exponent indicate the power of the quantity, and the denominator the root of that power. (Art. 71.) The square of ax is a2x2. We square both factors, and so, for any other powers, we raise all the factors to the required power. Conversely, then, we extract roots by taking the required roots of all the factors. Thus the cube root of 8x3 is 2x. The cube root of the factor 8 is 2, and of the factor x is x. The cube root of 16a3 cannot be expressed in a rational quantity, but it can be separated into factors, 8a3× 2, and the cube root of the first factor can be taken, and the index of the root put over the other factor thus, 2aX2, or 2a3/2. In such cases, the radical sign is usually preferred to the fractional index, as making a more distinct separation between the factors. The square root of 64a is obviously 8a3, and from this and the preceding examples we draw the following RULE. For the extraction of the roots of monomials. Extract the root of the numeral coefficients and divide the exponent of each letter by the index of the root. EXAMPLES. 1. What is the square root of 49a2x2? 2. What is the square root of 25c12? 3. What is the square root of 20ax? Ans. Tax2. Ans. 5cb. Ans. 2/5ax. In 20, the square factor 4 can be taken out; the other factor is 5. The square root of 4 is 2, which is all the root we can take; the root of the other factors can only be indicated as in the answer. 4. What is the square root of 12a2? Ans. 2a√3. 5. What is the square root of 144a2c2x2y2? 6. What is the square root of 36xa? Ans. 12ac2xy. Ans. 6x2. (Art. 72.) The square root of algebraic quantities may be taken with the double sign, as indicating either plus or minus, for either quantity will give the same square, and we may not know which of them produced the power. For example, the square root of 16 may be either +4 or -4, for either of them, when multiplied by itself, will produce 16. The cube root of a plus quantity is always plus, and the cube root of a minus quantity is always minus. For +2a cubed gives 8a3, and -2a cubed gives 8a3, and a may represent any quantity whatever. 6. What is the 4th root of 256a1x? 7. What is the 4th root of 16a? s. What is the 4th root of 64x2y2? Ans. 4ax2. Ans. ±2a1. Ans. ±√8xy. N. B. The 4th root is the square root of the square root. N. B. Reduce the fraction as much as possible, and then ex To extract roots of compound quantities. (Art. 72.) We shall commence this investigation by confining our attention to square root, and the only principle to guide us is the law of formation of squares. The square of a+b is a2+ 2ab+b2. Now on the supposition that we do not know that the root of these terms is a+b, we are to find it or extract it out of the square a2+2ab+b2. We know that a2, the first term, must have been formed by the multiplication of a into itself, and the next term is 2axb. That |