CUBIC EQUATIONS. 1. BINOMIAL CUBIC EQUATIONS. (575.) A binomial cubic equation contains only the cube of the unknown quantity and a constant term." (576.) Let the equation be a3-a3=0 Assume hence xay, then a3y3 — a3 ... [1] or 33-1-0 y3 = 1, and y =1 It is evident that equation [2] is the product of the two factors (y-1) (y2+y+1)=0 also the roots of the factor y2+y+1=0 1 1 are (277) −2+2√3, and 2 2 (577.) Denoting these two quantities respectively by a and 8, it is evident that equation [3] is verified by substituting for y any one of the three quantities 1, a, ß; hence these are the three roots of the equation [2]; and the roots of [1] are therefore x = a, as, aß. 1 1 (578.) If a′ and s' denote 1⁄2 +√3, and 2 2N respectively, it may be similarly shown that the three roots of the equation y2+1=0 are 1, a', '; and that the roots of (579.) Hence 1 has three cube roots, namely, 1, a, ß; and I has also the three cube roots 1. Given a 8, to find the three values of x. = Here a38, and a=2; hence the three roots a, aa, aß, are 2, 2 a, 2 ß. 2. Given 23=-5, to find the values of x. Here a35; and hence [4] the roots-a, aa', as', become 5, a'/5, 6'/5, where 5 1-702976. - EXERCISES. 1. Given 23 = 64, to find the values of x. 2. COMPLETE CUBIC EQUATIONS. (580.) Let the complete cubic equation y3+py2+qy+r=0 be given to find its roots. 1 ... (581.) Let y=x-p, and find from this the values of 33 y3 and y2, and substitute in the above equation for y, y2, and y3, their values, and it will be reduced to the form the three roots of the equation [2] are 1 27 1 2723)} x=u+v, au+ßv, and ßu + uv It is only the first of these roots that is usually given, or The value of a being known, that of y is easily found (581), (583.) Given the equation y3-9 y2+ 25 y—25, to find the values of y. Comparing this equation with [1], p=-9; hence let ya+3 by (581) =- - 2, Again, comparing this equation with [2], p = 8 N I 1 92 27 p3)} 8 27 = ~ {2+√(4 — — ) } +√ {2—√(4— } =(2+1.9245) +√(2−1·9245) = /3·9245+/0755 = 1.5773+4227=2 Hence y=x+3=2+3=5 The other two roots of x are (577) 2 ∞, 26, and those of y are 3+2a, 3+2ß 1. Given 23. 3x 2. Given 23. EXERCISES. 180, to find x. 6x+9=0, to find x. 3. Given 23—8 x2 + 15 x — 18=0, to find x. 4. Given a3. · 15 x2+75 x —1330, to find x. 5. Given 3+ 15 x2 + 85 x + 1750, to find x. ANSWERS. 1... x=3 4... x=7 7... x = (50+5 √235) + √ (50 — 5 √ 235) This last value of x in the 7th exercise the student may reduce to a single number by extracting the roots. (584.) In order to derive the formula given in art. (582), let there be given the equation Assume two unknown quantities, u and v, such that x=u+v, then [1] becomes u3 +v3+(3 uv+p) (u+v)+q=0 ... [2] (585.) As u and may have an indefinite number of systems of values to satisfy the equation x = u+v, another condition is necessary to make their values determinate (263); hence assume u3+v3+q=0 and then it follows from [2] that [3] (3 uv +p) (u + v) = 0, or 3 uv +p=0, or 3 uv=-p then substituting this value of v in [3], and reducing, it becomes (586.) This equation, called the reduced equation, being of a quadratic form (284), may be solved as a quadratic, and thus the two values of us are easily found to be - 1 1 -9±√(2+23) But 3 would be found to have 27 203) exactly the same values, as u and v are similarly involved; hence one of these values must be assumed for u3, and the other for 3; therefore u+v. and hence But 3 has three values (577), u, au, ßu, and 3/23 has also three values, v, av, Bo; hence, since the values of x are the sums of any two of the values of u and v, and since there are 3 × 39 pairs of values of these two quantities, there are apparently as many values of x. The condition, however, in art. (585) that 3 uv+p=0, or 3 uv =— -p, requires that the product of the values of u and v should be a possible or real quantity, namely, -p; and this circumstance excludes six of the nine pairs of values of u and r, so that the only admissible values are thus found to be x=u+v, x=aa+ßb, x=ßa + ab (587.) When the given equation is a complete cubic, its second term may be taken away by the method given in art. (581), and then it assumes the same form as the equation [1] in art. (584.) (588.) An equation of the form a31 +pa2n +qx2+r=0 may be reduced to a cubic, by assuming az, when it becomes 23+ pz2 + qz+r=0. 1 1 When p is negative, and p3 > 27 42, the values of a given by the preceding formulas are all evidently imaginary; and yet it is known by the theory of equations, or even by developing the values of x by the binomial theorem, that they are in this case all real. This case is called the irreducible case, because in it the value of a cannot generally be found in finite and real terms; and, in fact, a value of x can generally be found, by known methods, in finite terms only when two of the roots of the equation are imaginary. The cause of this paradoxical analytical fact is, that the equations in articles [3] and [4], namely, |