Of Cubic Equations. 95. A cubic equation is that in which the unknown quantity rises to 3 dimensions; and, like quadratics, or those of the higher orders, is either simple or compound. A simple or pure cubic equation is of the form b ax3-b, or x3--; where x = a A compound cubic equation is of the form x2+ax=b, x3+ax2=b, or x3+ax2 + bx=c, b in each of which the known quantities a, b, c, may be either + or -. Or either of the two latter of these equations may be reduced to the same form as the first, by taking away its second term ; which is done as follows: RULE. Take some new unknown quantity, and subjoin to it a third part of the coefficient of the second term of the equation with its sign changed; then, if this sum, or difference, as it may happen to be, be substituted for the original unknown quantity and its powers in the proposed equation, there will arise an equation wanting its second term. Note. The second term of any of the higher orders of equations may also be exterminated in a similar manner, by substituting for the unknown quantity some other unknown quantity, and the 4th, 5th, &c. part of the coefficient of its second term, with the sign changed, according as the equation is of the 4th, 5th, &c. power. 1. Required to exterminate the second term of the equation x3+3ax2±b, or x3+3ax2-b=0. Here xz 3a 3 -a. -6-0; or z3-3a'z-b-2a3, in which equation the second pow er (22) of the unknown quantity is wanting. 2. Let the equation 23-12x2+3x-16 be transformed into another that shall want the second term. Here (z+4)3=z3+1222+48z+ 64 -12(x-4)=-1222-96z-192 Then +2(z+4): = +3+12 z+4. Whence 245x-116--16; or z3-45z-100, which is an equation where 22, or the second term, is wanting, as before. 3. Let the equation x3-622=10 be transformed into another that shall want the second term. Ans. 12z-26. 4 Let y3-15y+81y-243 be transformed into an equation that shall want the second term. Ans. x+6x-88. 5. Let the equation x+x+x-0 be transformed into another that shall want the second term. Ans. y3+1fx=4. Here y=x+5, y3=x3+15x2+ 75x+125 -15y— —15x2-150x-375 +81y= = -243= Here xy, +81x+405 -243 x+6x—88—0. Hence we have y3+1ty—=0. 6. Let x+8y3-5x2+10x-4-0 be transformed into another that shall want the second term. Ans. y'-29y+94y-92-0. Here xy-2, 29y2+94y 92 4 = 0. 7. Let x-3x3-3x2-5x-2-0 be transformed into another that shall want the third term. Now, the value of e, by which the third term is taken away, is had by resolving the quadratic equation 6e-9e+3=0; the roots of which are found 1, or hence, by substituting y+1 for x, in the given equation, we find y* + y2 4y-6-0, an equation wanting the third term. y=0; hence y3-2y+3=0; the roots of which are the reciprocals of the former. 9. Let x-x3-3x2-x's be transformed into another, in which the coefficient of the highest term shall be unity, and the remaining terms integers. Ans. y-3x+12y-162y+72—0. y3 y2 3y 1 + 64 2.63 3.62 4.6 18-y1—3y3 + 12y-162y+720, the equation required; the roots of which are 6 times those of the former. Of the solution of Cubic Equations. 97. RULE. Take away the second term of the equation when necessary, as directed in the preceding rule. Then, if the numeral coefficients of the given equation, or of that arising from the reduction above mentioned, be substituted for a and b in either of the following formulæ, the result will give one of the roots, as required.* * x3+ax=b. a3 T b2 a3 b b2 (1+27) or Where, it is to be observed that when the coefficient a, of the a as also in second term of the above equation, is negative, 27, as also the formula, will be negative; and if the absolute term b be negative, 6 in the formula will be negative; but 6 will be positive. It may likewise be remarked, that when the equation is of the a3 form x3-ax±6, and is greater than or 4a3 greater than 27 4' 2722, the solution of it cannot be obtained by the above rule; as the question, in this instance, falls under what is usually called the Irreducible Case of cubic equations. See index. 1. Given 2x-12x+36x-44, to find the value of x. Here 3-6x+18x-22, by dividing by 2; and, in order to exterminate the second term, put x=x+8=2+2: (z+2)=2+6x+12z+8 Then —6(z+2)2= +62-24z-24-22, 18(z+2)= 18z+36 *If, instead of the regular method of reducing a cubic equation of the general form +a+b+c=0, to another, wanting the second term, as pointed out in the preceding article, there be put x= ૐ(Y a), we shall have, by substitution and reduction, y3+(96—3a2)y=9ab-27c-2a3; where, since the value of y can be determined by either of the formulæ given in this rule, the value of x will also be known, being (y-a). And if b―0, or the original equation be of the following form, 3+ax2+c=0, the reduced equation will be 3-3a2y--27c, where the value of y being found as above, we shall have, as before, x=} (y—a), which formulæ, it may be observed, are more convenient, in some cases, than those resulting from the preceding article; as the coefficients thus obtained are always integers; whereas, by the former method, they are frequently fractions. Whence 2+6x+20-22, or z2+6z=2; and, consequently, by substituting 6 for a and 2 for b, in the first formula, we shall have X = ~ { ? + √({ + 22)} +~ {}−√(‡+31o)}= ✔{1+√(1+8)}+√{1−√(1+8)}=√(1+√9)+ = .~/(1 —√/9)=~/(1+3)+(1–3)=√/4—3/2; therefore z +2==√4√2+22+1.587401-1.259921-2.32748, 2. Given x3-6x=12, to find the value of x. Ans. Here a being equal to -6, and b equal to 12, we shall have by the formula z={6+(36—8)}— -2 ✔{6+√(36—8}} 2 (6+5.2915)+ (6+5,2915) .8957-1392; therefore x=3.1392, the answer. 3. Given x3-2x-4, to find the value of x. Here a being =-2, and b-4, we shall have, by the forx=~{−2+√(4—27)}+~{—2—/(4)}, or by reduction, mula, (2+3)-(2/3)=(-2+1.9245) -(2+1.9245)=(-.0755-3.9245-1226-1.5773-1.9999, or -2, Ans. Note. When one of the roots of a cubic equation has been found, by the common formula as above, or in any other way, the other two roots may be determined as follows: Let the known root be denoted by r, and put all the terms of the equation, when brought to the left hand side, =0; then, if the equation so formed be divided by xr, according as r is positive or negative, there will arise a quadratic equation, the roots of which will be the other two roots of the given cubic. 4. Given x3-15x=4, to find the three roots or values of x. Here is readily found, by a few trials, to be equal to 4, and therefore, by division, we get x-4)x3—15x-4(x2+4x+1; Whence, according to the note above given, x+4x+1=0, or x+4x=-1; the two roots of which quadratic are -2+No3, and-2-3; and consequently 4, -2+/3, and -2-3, are the three roots of the proposed equation. 1. Given 2+3x2-6x=8, to find the root of the equation, or the value of x. See index. = Ans. x 2. 2. Given x3-6x=6, or x3+9x=6, or 23+2x2-23x = 70, or 2-17x+54x-350, or x-3x-5, or 500, to find the value of x in each expression. Answers, 4+2, or 9—3, or 5.1346, or 14.954, or 3. 426, and 7.61728, as required. 3. Given x3-6x-4, to find the three roots of the equation, or the three values of x. Ans. -2, 1+/3, and 1-3. 4. Given x3-5x2 + 2x =— - 12, to find the three roots of the equation, or values of x. Ans. 3, 15, and 1-5. 6. Find the root of the equation x3 + x =500, or x3-6x-5.6. Ans. x 7.89500828, and x=2.82526384. Solution of Cubic Equations by the Tables of Sines and Tangents. 98. Cubic Equations may often be more readily resolved by means of the following trigonometrical formulæ, than by the method before given; particularly when they fall under what is usually called the irreducible case, in which the common method is known to fail. 1. Given +ax=b, to find the value of x. 2ar a Put √1⁄2= tan. z, and √(r2 tan. §z)= tan. u, 36 2 a Then = cot. 2u; where the upper or under sign is to be taken according as b is positive or negative. 2. Given x2-ax=b, to find the value of x. This form has two cases, according as 2ar a 3 36 is less greater than radius r, or 4a3 less or greater than 2762. In the first of these cases, or when 4a3>2762, Put 2ar 36 a sin. z, and (r2 tan. §z)= tan. u; No3 2 a or 30 Then xx cosec. 2u, where the upper or under sign is 3 to be taken, according as b is positive or negative. In the second case, or when 4a3<27b2, 3br 3 Put N cos. z, then x will have the three following values; 2a a where the upper sign is to be used when b is positive, and the under sign when it is negative, as in the other cases. 1. Given +18x=6, to find the only real value of x. Here a 18, and b-6; whence 2ar a and, consequently, by formula 1, we shall have |