x(1 + 3 8 √2 1), (10), it is clear, if we develop the binomials in the right members of these equations, that we shall get the values of the first members of the equations by converging series when and are fractional. 4 n 8 n 1 Thus, if n = 3, and we put for r, and for a in (8), we shall get (1+1=1)2= 1 2 = 0.94563651 + 0.67252751 -1; and by putting for a in (8), we shall also have (1 3 = 0.97555394 + 0.44623019 -1; consequently, from (9) Changing 1 into +1, this result clearly indicates that two of the roots of the equation 3+1=0 are 0.5 ± in Ex. 4 of Recurring Equations. To find the remaining root of a + 1 = 0, we shall put n = 3 and m = 3 in (6), and shall get x = 1, which also agrees with the answer given in the example previously cited. It is clear from (10), that if we square the preceding roots, we shall get 1±√3 20-1=0. 2 and 1 for the roots of the equation Hence the method of proceeding in order to find the roots of any binomial equation becomes too evident to require any further comment. (6.) Finally, we have shown, from the principles of pure algebra, that any binomial equation must have as many different roots as there are units in the index of the unknown letter. We have also given a simple method for finding the roots of any such equation to any required degree of exactness, so that we may consider the solution of binomial equations-as being completed. SECTION XVII. SOLUTION OF CUBIC AND BIQUADRATIC EQUATIONS. (1.) IN conformity to what has been shown, we may clearly take y+A1y2+ Ay = As, (1) for the general representative of any cubic equation. If for y we put x - Α. A, it is easy to perceive that (1) will 3 be reduced to the form a + ax = b, (2), which is of a much more simple form than (1). Putting xp+q, and substituting p + q for x in (2), it becomes (p + q)3 + a(p + q) = b, or p3 + q3 + (3pq + a)(p + q) = b; consequently, by assuming 3pq + a = 0 or pq = α (3), we get p3 + q3 = b, (4). From (4) and (3) we easily get (p3 + q3)3 — 4p3q3 = b2 + The half sum and half difference of (4) and (5) give p3 : b + V 2 4 27 3 b + and q= V 2 a3 = 2 4+27; consequently, by taking the sum of the cube roots of these expressions (since If a is positive, while b is positive or negative, it is clear that can be found from (A) or (A'); also, if a is negative 2 3 it is clear that a3 4 27 will be imaginary; consequently, a can not be found from (A) or (A'), unless the cube roots can be extracted in such a way that the imaginary parts shall destroy each other; and any example in which the cube roots can not be thus extracted is said to fall under the irreducible case of cubic equations. To illustrate (A) and (A'), take the following EXAMPLES. 1. To find a from the equation a3+3x=2. x Since a and b are here expressed by 3 and 2, we easily get from (A) and (A'),∞ = (1 + √2) + (1 − √2) and a = 2. To find a from the equation a3-6x=6. By putting a + 1 for y, the equation is easily reduced to a3-3x=7. Solving this equation by (A'), we get x = to the value of x, we get the value of y. Thus we perceive how to apply (A) or (A') to the solution of any cubic equation which does not fall under the irreducible case. (2.) SOLUTION OF CUBIC EQUATIONS WHICH FALL UNDER THE IRREDUCIBLE CASE. a for a in (2), it becomes as Putting ax = b, (1'); in which may be positive or negative, while a must be positive, and at the same time must not be greater than max + max = 0 to the first member of (1), it тах + тах max= 0, or a2 ax = b; consequently, if we ma, we shall get max ax= ing these values of a2, we shall get m(m − 1)2 = 2)x(+1)=0, the roots of this equation a3 — 27' for v in (2), we get x = 2 3' Substituting 3' for the a3 N, and assuming V3 (2"); then it is clear that we may consider (3') as being derived from (1"), by the same process that (3') has been derived from (1'); and it is evident, if we can find V from (1"), that (2") will give the value of v, and then (2′) will give the value of x. In like manner (1") may be supposed to be derived from (3) may be supposed to be derived from s a3 that N is not greater than 2; and if N is less than 2, it is clear that N+ 2 is less than 2, and nearer to it than N; and (N +2 + 2) is also less than 2, and nearer to it than N+2; and so on to infinity. Hence, we may manifestly consider the irreducible case |