It is easy to see that is the sought product, so that + Ox3y + Oxy2 + y3 = x2 + y3 we have (2xy + y2) (x + y) =a+y; consequently, xy + y2 and x + y are factors of a3 + y3. It is clear that if we change y into y in the last example, the multiplicand and multiplier there given, will become the multiplicand and multiplier in this example; also, the results there obtained will coincide with those of this question. Ex. 8. To develop (a + x2y + xy3 + y3) (x − y), by detaching the coefficients. Detaching the coefficients, etc., we shall have 1 + 1 + 1 + 1 1-1 1 + 1 + 1 + 1 -1-1-1-1 1+0+0+0−1 the sum of the products of the coefficients. - Supplying the requisite letters, we shall get (3 + x2y + xy2 + y3) (x − y) = x1 — y1, as required; of course, ∞3 + x2y + xy2+ y3 and x y are factors of 1-y1. Since is a factor of a3 + x2y, and y2 a factor of xy + y3, we have ∞3 + x2y + xy2 + y3 = x2 (x + y) + y2 (x + y) = (x2 + y2) (x + y); consequently, (3 + x2y + xy2 + y3) (x − y) reduces to (~2 + y3) (x + y) (x − y) = (x2 + y2) (.x2 — y2) = x1 — y1, as it clearly ought to do; for x1 — y1 = (x2)2 — (y2)2 = (x2 + y2) (x2 — y3), as appears from Ex. 5. Conversely, having a -y to find its factors; — x3y + x3y — x2y2 + x2y2 — xy3 + xy3 = 0, or = 0 to it, and we get x - y = x1 — x3y + x3y we may add — xy3 + xy3 — y1 = x3 (x − y) + x2y (x − y) + xy2 (x − y) + y3 (x − y) = (x3 + x2y + xy2 + y3) (x − y), or x1 — yk = 20k — x2y2 + x2y2 — y' = x2 (x2 — y3) + y2 (x2 — y2) = (x2 + y2) (x2 y3) = (x2 + y2) (x + y) (x − y); consequently, x + x2y + xy3 + y3 and x y are factors of y; and so are x2 + y2 and x2 — y2, or x2 + y2, x + y, and x y. REMARKS. If we have a"-b", such that n is any positive integral number; then if we put b for a, we have b" - 6" = 0; there fore, ab must be a factor of a"-b", agreeably to the remarks under Ex. 4. To find the factors of a"-b", we may evidently put it under the form a"-b" = an an-1b + an-1f — an −272 + an-an-373 + an-373 — etc.,. = - 1 -1 ... -ab"-1+abr-1_ b” — a” — 1 (a — b) + an − 2b (a — b) + an − 3b2 (a — b) + an−173 (a - b) + etc.,. . . . +abn−2 (a - b) + br−i (a - b) = [an-1 +an-2b+ a"−372 + a2-473 + etc., . . . . + ab” - 2 + br−1] [a — b], (A); consequently, a"-b" is resolved into the two factors a"-1+ an−2b + an−372 + an―473 + etc., + abn - 2 +b-1, and ab; and it is evident that the first of these factors has as many terms as there are units in n. We have written (A) against the equation that shows a"-b" to be the product of the factors found, and shall refer to the equation when necessary as equation (A), or formula (A); and we shall adopt a similar method of designation in all analogous cases, whenever it shall be of importance to do it. If n is an odd number, and we put b for b in equation (A), it will evidently become a" + b′′ = [a”-1 [an-an-2b+ an-373-an-473 + etc.,... ab"-2 + bn-1]. [a + b], (B), as is clear from the consideration that any odd power of a negative quantity is negative, and that any even power of a · negative quantity is positive. Hence we see that an-ban has the factors in the right members of (A), and (B) for its factors; consequently, a2 ban has (a - b) (a + b) = a2 — b2 for a factor. If in (A) we put n = 2, it becomes ab= [a2¬1 + a2-2b]. [ab], or since 2-1 = 1, and 2-2 = 0, we have a2b2 = [a + ab]. [a b], or (a + b) (a − b) = (a + aob) (a - b); of course, we must have a + ab = a + b, or ao = 1. Hence, since a may represent any quantity, it follows that any quantity whose exponent is 0 is unity (or 1). = = Again, if in (A) we put n = 3, and observe that a3-3 a° =1, we shall get a3-b3 (a2 + ab + b2) (a - b); and in like manner, if n = 4, we get a b1 = (a3 + a2b + ab2 + b3) (a - b); also, a3 — b3 = (a1 + a3b + a2b2 + ab3 + b1) (a — b) ; and so on, for any particular, positive integral numbers, that may be assumed for n. There is a transformation of (A) which will be useful to us in the investigation of the binomial theorem, which we shall now proceed to give. an Thus, since a"-b" is less than a" by b", if we add b" to the right member of (A), we shall clearly have a"=b" + [an-1 + an-2b + an¬372 + an−463 + etc., . ... +abn―2 + b-1]. [ab]; and if we put a-bh, by adding b to these equals (Ax. II.), we get a − b + b = b + hor (since -b+b=0, Ax. VI.) a = b + h; consequently, by putting h for ab and b+h for a, in the preceding equation, it will become (b+ h)" = b" + [ (b + h)n-1 + (b + h)"−2b + (b + h)n-372 + (b + h)-463+ etc.,..... + (b + h)bn-2 + br-1]. h, (C), which is the transformation required. It will be observed that b+ h is a binomial quantity, since it consists of two terms, b and h, united by the sign +, according to 14 of Sec. I.; and that (b + h)" denotes the power of b+h, whose exponent is n, agreeably to 9,and 15, of Sec. I. - 3, n The right member of (C) is one form of the development of (b + h)", which has the term b" that is independent of h, and its remaining terms (or the terms within the braces, [],) dependent on h, since they have h for a common factor. Since the exponents n − 1, n — 2, n - 4, and so on, of bh within the braces in (C) are positive integers, we may develop (b+ h)"−1, (b + h)"-2, (b + h)-3, and so on by changing into n-1, n-2, n-3, and so on; and we shall have (b + h)"−1 = bn-1 + terms that depend on h; also (b + h)n2 = 7n−2+ terms that depend on h, and (b + h)”—3 = b-3+ terms that depend on h, and so on to b + h. -1 Hence, since n−2b = bn−2+1 = Zn−1, Un−372= &n−3+2 = fn−1, and so on to bb"-2 = br-1, we get (b + h)n−1 + (b + h)n−2b + (b + h)n−3b2 + (b + h)n-473 + etc.,.... + (b + h)b" —2 + bu−1 == h)n−473 br−1 + Zn−1 + Zn−1 +, etc., to n terms, + terms that depend on h, nb"-1+ terms that depend on h; consequently, putting for the terms within the braces in (C) their value nỗ"-1 + terms that depend on h, we get (b + h)" = b2 + (nʊn−1 + terms that depend on h,). h, or multiplying the terms in the parenthesis by h, we get (b + h)” = b" + nb”−1h + terms that depend on h2, h3, etc. Ex. 9. To develop the square of x+a; or, which is the same, to find the product arising from the multiplication of x + a by x + a2 which is expressed by (x + a) (x + a) = (x + a)2. = ། Here we have, by using the common method of multiplication, We shall now find the square of a by detaching the coefficients. The coefficients of a + a being detached give 1 + 1, so that we have to multiply 1 + 1 by 1+1. Hence we get 1+ 1 1+ 1 1+1 1+2+1 = the sum of the products of the coefficients, and supplying the letters that correspond to the coefficients in the product, we get a2 + 2xa + a2 = (x + a); which is the same result that we found by the common method of multiplication. Hence, the square of the sum of any two numbers or quantities of the same kind, is greater than the sum of their squares by twice their product. For if x is one of the numbers or quantities, and a the other, then a + a is the expression for their sum; and a2 + a2 is the sum of the squares of x and a, and twice the product of x and a is 2xa; and if we add 2xa to x2 + a2, we get x2 + a2 + 2xa = x2 + 2xa + a2 = (x + a)2 = the square of +a, as we have shown. It is clear that the method of squaring a number which consists of more than one digit (arithmetically) is very similar to the method of squaring x + a. Thus, to square 23, we have 4 23 23 69 46 = the product of 23 by 23, or 232; where it will be observed that in multiplying 3 by 3 we get 9 = 32, and in multiplying the 2 by 3 we get 6, which is really 60, on account of the local value of 6; also in multiplying 3 by 2 of the multiplier we get 6, which is 60, on account of the local value of the 2; and multiplying the 2 of the multiplicand by the 2 of the multiplier we get 4= 22; or, on account of the local values of the 2 in the multiplicand and multiplier, the 4 is really 400. Hence, the square of 23 is equal to the sum of 400, 60 × 2, and 9, which is 400 + 120 + 9 529, which is the same as x2 + 2xa + a2 when we put 20 for x and 3 for a. In like manner it may be shown, if we square any number, that the process of squaring is the same as to separate the number into any two parts, and to add together the squares of the parts and twice their product. Thus, if we take 7896, it may be separated into the parts 7000 + 896, 7800 + 96, or 7890+ 6; and we shall have 78962 = 70002 + 2 × 7000 × 896 + 8962 = 78002 + 2 × 7800 × 96 + 962 = 78902 + 2 × 7890 × 6 + 62 = 62346816. Ex. 10. To develop (x - a) = (x − a) (x − a), or to find the square or second power of a a. - a by itself, we get Here, by multiplying — a x a = a will (a) as required. Hence, if a and a represent any two numbers or quantities of the same kind, then a represent their difference, and x2 + a2 will express the sum of the squares of x and a, and 2xa represents twice the product of x and a; consequently, since (a) = x2 — 2xa + a2= x2 + a2 - 2xa, it follows, that the square of the differ ence of any two numbers or quantities of the same kind, is less than the sum of their squares by twice their product. Thus, since 1920-1, we have 192 202 +1° - 2 x 20 =401 - 40 = 361; also, since 397 400 — 3, we get 397 =4002 + 32 - 2 x 400 x 3 = 160009 — 2400 = 157609. = |