The Cubic Formula

                                                                                 David W. Hansen
                                                                                        © 2003   

 

               The general cubic equation is ax3 + bx2 + cx + d = 0. Dividing both sides by a, it can be simplified to

        x3 + px2 + qx + r = 0. By a simple substitution, which is shown below, the quadratic term may be

        eliminated, leaving us with x3 + mx = n.

               Scipione del Ferro (about 1515) is reputed to have been the first to have discovered a formula for

        solving   x3 + mx = n. [1]    As given by Cardano in his Ars Magna, the proof is as follows.

               We know that                                    (a - b)3    =    a3  -  3a2b  +  3ab2  -  b3,

        or                                                (a - b)3  +  3a2b  -  3ab2    =    a3   -   b3,

        and                                               (a - b)3  +   3ab (a - b)      =    a3   -    b3.                                                    (1)

                Letting x  =  a - b   in (1) above, we have    x3  +  3abx  =  a3  -  b3,   or    x3  +  mx  =  n,
        where                          m  =  3ab      (2)            and             n  =  a3  -  b3.                                                       (3)
        To find a, we solve for b in (2) to get   b = m / (3a). Substituting this value of b into (3), we get  
                                         n = a3 -  [ m / (3a) ]3,             or              n  =  a3  -  m3 / (27a3).                                     (4)
         Multiplying (4) by 27a3, we have                 27na3  =  27a6 -  m3,
         or                                           27a6  -  27na3  -  m3   =   0,                           which is a quadratic in a3.
         Letting  y  =  a3  gives us     27y2   -  27ny   -   m3   =   0.                     By the quadratic formula,
                                      _______________                                                           _______________________
      y    =   [ 27n +/ - Ö (27n)2 + 4(27)m3    ] / 2(27)     =   27n / 2(27)   +/ -  Ö [ (27n)2 + 4(27)m3 ] / 4(272)
                    
                                                                             ______________________________
                                      y    =    (n / 2)     +/ -  Ö   272n2 / 4(272) +   4(27)m3 / 4(272)     
                                                                              _________________
                                      y    =    (n / 2)     +/ -    Ö   n2 / 4 +   m3 / 27    
                                                                              _________________
                                      y    =    (n / 2)     +/ -    Ö   (n / 2)2 +   (m / 3)3 .             Replacing  y  by  a3,   we have
                                                                               _________________
                                     a3   =    (n / 2)     +/ -    Ö   (n / 2)2 +   (m / 3)3                                                                
 
                                                  3  _______________________________
                                      a    =   Ö (n / 2)     +/ -   [ (m / 3)3 +   (n / 2)2  ]1/2    .                                                           (5)
 
                  To find b, we now solve for a in (2) above and substitute its value into (3) which,  after solving
         for b by the same procedure as for a, gives us
   
                                                   3  _______________________________
                                        b    =   Ö - (n / 2)     +/ -   [ (m / 3)3 +   (n / 2)2  ]1/2    .                                                       (6)
 
         Thus, the solution to   x3 + mx = n   is   x = a - b, where a and b are given by equations (5) and (6).
 
         Example 1.   Use the cubic formula to solve   x3  +  63x  =  316.    Using formulas (5) and (6) above
         with m  =  63 and n  =  316, we have  m/3  =  21,  n/2  = 158,  and 
                                                                             __________                            _____________  
                                             a3   =  158   +/ -   Ö 213 + 1582     =   158  +/-   Ö 9261 + 24964   
                                                                              ______
                                                    =  158   +/ -   Ö34225               =   158  +/ - 185.
 
          Using the  +  root,              a3  =  158  +  185   =   343,         and   a   =    7.                                              (7a)
 
          Using the   -  root,              a3  =  158   -  185   =   - 27,         and   a   =  - 3.                                              (8a)
                                                                               _________                               _____________
          Also,                          b3   =  - 158  +/ -   Ö 213 + 1582     =   - 158  +/ -  Ö  9261 + 24964  
                                                                              ______
                                                    =  - 158  +/ -  Ö 34225             =    - 158  +/ -  185.
 
         Using the  +  root,              b3  =  - 158  +  185   =     27,          and   b   =   3,                                             (7b)
 
         Using the   -  root,              b3  =  - 158   -  185   =  - 343,          and   b  = - 7.                                            (8b)
 
         Since  x  =  a  -  b,  using (7a) and (7b),  we have x   =   7  -  3   =   4.
 
         Using (8a) and (8b), we have x  =  ( - 3)  -  ( - 7)   =   4  again.     Thus, we can eliminate the negative
         root in formulas (5) and (6) for a and b above giving us
             ___________________________________________________________________
                 The Cubic Formula  
                                                                                                              
                   If   x3  +  mx  =  n,  then  x  =  a  -  b,  where   a3  =   (n / 2) + D,   b3  =  - (n / 2) + D 
                                            _______________
                  and       D =   Ö (m / 3)3 +   (n / 2)2   .                                              
               __________________________________________________________________
 
 
         Example 2.  Solve x3 + 6x = 7. Since m = 6 and n = 7, we have
                                                        _____________             _________               _____
                                            D  =  Ö (6/ 3)3 + (7/ 2)2      =    Ö 8 + (49 / 4)      =    Ö 81 / 4   =   9 / 2.
 
                                            a3  =     7/2  +  9/2  =  16/2  =  8,     and     a = 2.
 
                                            b3  =   - 7/2  +  9/2  =  1,      and     b = 1.     Thus,   x = a - b   =   2 - 1 =  1. 
    
         Example 3.  Solve x3 + 6x = - 7. With  m  =  6  and  n  =  - 7,  we have
                                                         _______________            ___________           _____
                                            D  =  Ö (6/ 3)3 + ( - 7/ 2)2      =    Ö 8 + (49 / 4)      =    Ö 81 / 4   =   9 / 2.
 
                                            a3  =   - 7/2  +  9/2  =  2/2  =  1,        and     a = 1.
 
                                            b3  =     7/2  +  9/2  =   8,        and     b =   2.     Thus,  x  =  a  -  b  =  1 - 2  =  - 1.     
 
 
        Example 4.   Solve x3 - 3x = 2.  With  m  =  - 3  and  n  =  2,  we have
                                                         ______________              _______             __
                                            D   =  Ö ( - 3/ 3)3 + (2/ 2)2      =    Ö ( -1) + 1     =    Ö 0   =   0.
 
                                            a3  =      2/2  +  0  =   1,       and     a = 1.
 
                                            b3  =    - 2/2  +  0  = - 1,      and     b = - 1.    Thus, x  =  a  -  b  =  1  -  ( -1 )  =  2.      
 
 
         Example 5.  Solve x3 - 6x = - 9.  With m  =  - 6  and  n  =  - 9,  we have
                                                    ________________             ___________            _____
                                       D   =  Ö ( - 6/ 3)3 + ( - 9/ 2)2      =    Ö - 8 + (81 / 4)     =    Ö 49 / 4   =   7 / 2.
 
                                       a3  =   - 9/2  +  7/2  =  - 1,     and     a = - 1.
 
                                       b3  =     9/2  +  7/2  =     8,    and     b =    2.    Thus, x  =  a  -  b   =   - 1  -  2   =   - 3.      
 
 
         Example 6.  Solve  8x3 + 168x + 279 = 0.      First, we must put this equation in the standard form of
         x3 + mx = n. Thus, dividing the equation by 8 and moving the constant term to the right-hand side, we
         get   x3  +  21x  =  - 279 / 8.   Then, m  =  21,  n  =  -  279 / 8, and we have 
                                                     __________________             _________________
                                         D  =  Ö (21/ 3)3 + ( - 279 / 16)2     =   Ö 343 + (77,841 / 256)
                                                     _________________________            ____________
                                               = Ö (87,808 / 256) + (77,841 / 256)     =   Ö(165,649 / 256)     =    407 / 16.
      
         So,                          a3  =   - 279/16  +  407/16   =  128/16     =     8,          and     a =     2.
 
                                         b3  =     279/16  +  407/16   =   686/16    =   343/8,     and     b =   7/2.
          Thus,  x  =  a  -  b   =    2  -  7/2   =   4/2  -  7/2  =   - 3/2.   
               Now let’s solve a cubic equation with an x2 term. Consider x3 + px2 + qx + r = 0. To eliminate the
         x2 term, we will let x = y - h, getting
                                                             (y - h)3  +  p(y - h)2  +  q(y - h)  +  r  =  0,
                                              y3 - 3y2h + 3yh2 - h3 + py2 - 2pyh + ph2 + qy - qh + r  =  0,
                                       y3 + (- 3h + p) y2  + (3h2 - 2ph + q) y + ( - h3 + ph2 - qh + r)  =  0.
 
         This last equation is a cubic equation in y. If we wish to eliminate the y2 term, we simply make its
          coefficient zero.  Thus, - 3h + p = 0,  so  h = p / 3. Then,  x  =  y - h  =  y - p / 3  and we have
                _____________________________________________________________
                          
                       To eliminate the x2 term from the cubic equation x3  + px2 + qx + r = 0,
                       simply replace   x   by    y - p / 3,   and simplify the equation.
               _____________________________________________________________
         Example 7.  Solve  2x3  +  24x2  +  114x  +  148  =  0.  We first divide by 2, the coefficient of
         the x3 term,  to put the equation in the standard form  x3  +  px2  +  qx  +  r  =  0.  This gives us
         x3  +  12x2  +  57x  +  74  =  0.   Then,  p  =  12,  and to eliminate the x2 term, we replace x
         by   y  -  p / 3  =  y - 4.                               So,  x  =  y  –  4.                                                                               (9)
        This gives us                            (y - 4)3  +  12(y - 4)2  +  57(y - 4)  +  74  =  0,
                                                y3 - 12y2 + 48y - 64 + 12y2 - 96y + 192 + 57y - 228 + 74 = 0,
         and               y3 + 9y - 26  =  0,     or      y3 + 9y  =  26,   which is in the form  y3  +  my  =  n.
         Using the cubic formula with m = 9 and n = 26, we get 
                                               ______________             _______              ________         ___ 
                                 D   =   Ö (9/ 3)3 + (26/ 2)2      =    Ö 33 + 132     =   Ö 27 + 169    =  Ö196  =  14 .
 
                                 a3   =      26/2 + 14   =   13 + 14  =  27       and     a  =  3.
 
                                 b3   =    - 26/2 + 14   =  -13 + 14  =    1       and     b  =  1.
          Thus,  y  =  a  -  b  =  3  -  1  =  2,  and from  (9)  above,  x  =  y  -  4  =  2  -  4  =   - 2.
       
                 And now we come to the most challenging types of cubic equations; namely, those which involve
         finding cube roots of complex numbers to obtain a solution.
      
         Example 8.  Solve  x3 - 21x = - 20.  Here,  we have  m  =  - 21  and  n  =  - 20.     Then,
                               _________________           ____________         _________           ____              _    
                  D  =  Ö( - 21/ 3)3 + ( - 20/ 2)2    =   Ö( - 7)3 + ( - 10)2   =  Ö- 343 + 100    =  Ö- 243  =   9iÖ3,
                                                                      _                                      _
         and                           a3  = - 10 + 9i Ö3,            b3 = 10 + 9i Ö3.
 
                                            3 _________                3 ___________
         Thus,  x  =  a  -  b  =  Ö -10 + 9iÖ3       -         Ö 10 + 9iÖ3   ,   which is a decidedly complicated
         answer!      However,   (2 + iÖ3 )3   =  - 10 + 9iÖ 3,   and    ( - 2 + i Ö3 )3  =  10 + 9iÖ3. 
         Thus, a = 2 + iÖ3,      b = - 2 + iÖ3 ,  and   x  =  a  -  b  =  (2 + iÖ3 )  -  ( -2 + iÖ3 )  =  2 - ( - 2)  =   4 !
 
        ( To find these cube roots, we can use the method developed in  this article.  However, in general, If
           we can’t find the cube roots of a3 and b3, then we must leave our answer in terms of the difference
           between two cube roots. Not a nice result!)                                                                                                
                                                                                                                 
         Example 9.  Find a cubic equation which has a root of 2Ö2.   Then, solve this equation using the
         cubic formula to verify that 2Ö2 is indeed a root.   Since  x  =  a  -  b,  we must select values for
         a and b which make x equal to 2Ö2 . There are many choices for a and b;  let’s pick  a = 1 + Ö2 and
         b = 1 - Ö2 .  Then,  x  =  a  -  b  = (1 + Ö2 )  -  (1 - Ö2 )  =  2Ö2 .    To find m and n for   x3 +  mx  =  n,  
         formulas (2) and (3)  give us   m  =  3ab  and  n  =  a3 - b3.
         Now,                                a3  =  (1 + Ö2 )3   =   1 + 3Ö2 + 6 + 2Ö2      =  7 + 5Ö2 ,                                       (9) 
         and                                  b3  =  (1  - Ö2 )3   =   1  - 3Ö2 + 6  - 2Ö2      =   7 - 5Ö2.                                      (10)
         
         So,                             m  =     3ab      =  3(1 + Ö2 )(1 - Ö2 )   =   3(1 - 2)   =      - 3 ,     
         and                              n  =  a3  -  b3   =       (7 + 5Ö2 )  -  (7 - 5Ö2 )           =   10Ö2 .
                                 
         Thus,  x3 - 3x  = 10Ö2  is a cubic equation with a root of 2Ö2 . 
 
                  To solve this equation using the cubic formula, we have  m  =  - 3  and  n  =  10Ö2.         Then,
                           _________________              ____________             _______          __
                D  = Ö( - 3/ 3)3 + ( 10Ö2 / 2)2    =    Ö( - 1)3 + (5Ö2 )2      =   Ö - 1 + 50   =   Ö49  =  7,
                                        
                                   a3  =  5Ö2 + 7  =  7  +  5Ö2,      and      b3  =  - 5Ö2  +  7  =  7  -  5Ö2 .
                                                                                                                                   
         To find the cube roots of a3 and b3, note that (9) and (10) above tell us that a = 1 + Ö2  and  b = 1 - Ö2
         are the cube roots of a3 and b3 respectively.  Thus,  x  =  a  -  b  =  (1 + Ö2 )  -  (1 - Ö2 )  =  2Ö2,  as
         predicted.
 
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