Finding Cubes and Cube Roots of Numbers
                          of the Form a√b + c√d and R√b + S√d
 
                                                                            David W. Hansen
                                                                                   © 2008
                                                                                    _        _
          Consider any number N of the form N = a √b + c√d, where a, b, c, and d are integers. Then,
                                   _        _                _        _           _         _                              __                  _         _
                N3   =   (a√b + c√d )3  =  (a√b + c√d )2 (a√b + c√d )   =  (a2b + 2ac√bd + c2d) (a√b + c√d )
                                           _                _                 ___                ___                 _              _
                        =        a3b√b  +  a2bc√d  +  2a2c√b2d  +  2ac2√bd2  +  ac2d√b  +  c3d√d
                                          _                _                   _                   _                 _               _   
                        =       a3b√b  +  a2bc√d  +  2a2bc√d  +  2ac2d√b  +  ac2d√b  +  c3d√d
                                                                       _                 _                 _             _  
                                                       =     a3b√b + 3a2bc√d + 3ac2d√b + c3d√d
                                                                                           __                             __
                                                        =       (a3b + 3ac2d)√ b + (3a2bc + c3d)√ d . 
 
                                                        =      a( a2b + 3c2d )√b + c( 3a2b + c2d )√d                                              (1)
 
     This last expression (1) gives us a convenient way to find the cube of a√b + c√d. Here’s how.
 
                                                              Finding the cube of a√b + c√d
 
          To find the cube of a√b + c√d; that is, (a√b + c√d)3,
 
                    1. Write   a√b + c√d   in the form      a (                 )√b    +   c (                   )√d
 
                    2. Calculate a2b and c2d and put these values inside the parentheses in 1, like this:
 
                                                       a ( a2b    +     c2d )√b    +   c (   a2b      +       c2d )√d
           
                    3.   Multiply the second term in the first set of parentheses and the first term in the second
                        set of parentheses by 3 (the two inner terms), getting
 
                                                         a ( a2b    +   3c2d )√b   +   c ( 3a2b    +   c2d )√d
 
                 4. Add the numbers inside each of the two sets of parentheses.
 
                     5. Multiply the integers in front of the two radicals √b and √d to get the final answer.   
 
          For example,  let’s find the cube of  3√5 + 4√2  (where a = 3, b = 5, c = 4, and d = 2).
 
                      1. We write down       3 (                 )√5    +   4 (                   )√2
 
                      2. We calculate   a2b   =   9(5)   =   45   and   c2d   = 16(2)   =   32 and put these values
                          inside the parentheses in 1, like this:
 
                                                     3 (   45    +     32 )√5    +   4 (   45    +     32  )√2
          
                      3. We multiply the second term in the first set of parentheses by 3, and the first term
                           in the second set of parentheses by 3 (the two inner terms), getting
 
                                                 3 (  45    +    3(32)  )√5    +   4 (  3(45)      +       32  )√2
                               
                                                          3 ( 45   +   96  )√5   +   4 ( 135   +  32 )√2
 
 4. We add the numbers in the two sets of parentheses.
 
                                                   3(141)√5   +   4(167)√2
 
       5. Finally, we multiply the integers in front of the two radicals √5 and the √2 to get
 
                                                               423√5 + 668√2    =    (3√5 + 4√2)3.   
 
     Check:  (3√5 + 4√2)3   =   (3√5 + 4√2) (3√5 + 4√2) (3√5 + 4√2)
                                            =   (45 + 12√10 + 12√10 + 32)(3√5 + 4√2)
                                            =   (77 + 24√10)(3√5 + 4√2)
                                            =   (231√5 + 308√2 + 72√50 + 96√20)
                                            =    231√5 + 308√2 + 72(5√2) + 96(2√5)
                                            =    231√5 + 308√2 + 360√2 + 192√5   =  423√5 + 668√2.
                                                                                                           
                                                    Finding a cube root of R√b + S√d
                                                                                                                                     _         _
          Again, as we did above, let’s consider any number N of the form N = a √b + c√d, where a, b, c, and d
     are integers. Then,
                                                                 _         _                       ­_                 _                 _             _
                                      N3        =      (a √b + c√d )3    =    a3b√b + 3a2bc√d + 3ac2d√b + c3d√d
                                                                                   __                               __
                                                   =    (a3b + 3ac2d)√ b  +  (3a2bc + c3d)√ d . 
                                                                              __                       __
     Thus,                        N3         =                  R √ b        +        S √ d ,                                                                       (2)
 
     where                              R   =      a3b    + 3ac2d,                                                                                                 (3)
 
     and                                  S  =   3a2bc   +   c3d.                                                                                                    (4)
 
     Solving (4) for a2, we get                  a2   =   (S – c3d) / 3bc,                                                                              (5)
 
     and solving (3) for c2, we get            c2   =   (R – a3b) / 3ad.                                                                             (6)
         
          Now, if we take the cube root of both sides of (2), we get
                                                                3 _________________
                                                 N   =      √  R √ b   +   S √ d          =     a √b + c√d.
                                              __             __                          _         _      
     Thus, a cube root of R √ b   +   S √ d   is given by a √b + c√d,   where b and d are known from N, and we
     can determine a and c from (5) and (6) by trial and error. This gives us

 

 
            Cube Root of   R √ b   +   S √ d         
                                             __             __                     
              A cube root of R √ b   +   S √ d , where R, S, b, and d are integers is given, if it exists,
              by a√b + c√d, where b and d are known, and a and c are integers determined from
                                          a2   =   (S – c3d) / 3bc     and    c2   =   (R – a3b) / 3ad
              by trial and error.
             ___________________________________________________________________
      
   
     Examples. 
                                                      _             _
     1.  Find a cube root of 324√3 + 430√2.     Using the Cube Root Theorem above, we have   
                    _             _            _             _
          324√3 + 430√2  =   R√b   +   S√d,   so    R = 324,  b = 3,  S = 430,  and  d = 2.   Using these values
          to find a and c, we get                         a2   =   (S – c3d) / 3bc                
                                                                                =   (430 – 2c3)/ 9c.                                                                      (7)
   
          and                                                         c2   =    (R – a3b) / 3ad
                                                                                =    (324 – 3a3)/6a
                                                                                =    (108 – a3)/2a                                                                         (8)
                                                                                                                                                                                              
 
          To find a2, we shall substitute values of c (starting with 1) into (7) until we get a perfect square.
 
                                                                     c               a2 = (430 – 2c3)/ 9c
                                                            -----------------------------------------------------
          1                               428/9    = 47.55
          2                               414/18  =  23
          3                               376/27  =  13.9
          4                               302/36  =  8.39
          5                               180/45  =  4 (a perfect square)
 
          So, when c  =  5,  a2  =  4, and  a  =  2. Thus, a  =  2 and c  =  5.     Now, equation (8) must be satisfied
          also, so substituting our value for a into the right-hand side of (8), we get
         
                                                (108 – a3)/2a  =  (108 – 23)/2(2)  =  100/4  =  25  =  52  =  c2.
                                                                                                                 
          Thus, both (7) and (8) are satisfied by a = 2, c = 5, and
                                            _        _            _        _                                          _             _
                                       a√b + c√d   =  2√3 + 5√2   is a cube root of  324√3 + 430√2.
 
          Check:    (2√3 + 5√2)3   =  (2√3 + 5√2)( 2√3 + 5√2)( 2√3 + 5√2)
                                                   =   (12 + 10√6 + 10√6 + 50)(2√3 + 5√2)
                                                   =   (62 + 20√6)(2√3 + 5√2)
                                                   =   124√3 + 310√2 + 40√18 + 100√12
                                                   =   124√3 + 310√2 + 40(3√2) + 100(2√3)
                                                   =   124√3 + 310√2 + 120√2 + 200√3
                                                   =   324√3 + 430√2.
                                                                 _
     2. Find a cube root of  - 10   +   9i√3.     To use the Cube Root Theorem above, we rewrite   
                         _                    _        __               _             _
         - 10 + 9i√3   as   - 10√1 + 9√-3    =    R√b   +   S√d,   so    R  =  - 10,  b  =  1,   S  =  9,  and   d  =  - 3.  
 
          Using these values to find a and c, we get
 
                                                                             a2  =   (S – c3d) / 3bc                 
                                                                                   =    (9 + 3c3) / 3c.                                                                     (9)
   
          and                                                            c2   =   (R – a3b) / 3ad
                                                                                   =    (- 10 – a3) / (-9a)
                                                                                   =    (10 + a3) / 9a                                                                    (10)
 
          To find a2, we shall substitute values of c (starting with 1) into (9) until we get a perfect square.
 
                                                                     c                        a2 = (9 + 3c3)/ 3c
                                                                 -----------------------------------------------------------
         1                        12/3 = 4 (a perfect square)
 
          So, when c = 1, a2 = 4, and a = 2. Thus, a = 2 and c = 1.     Now, equation (10) must be satisfied also,
          so substituting our value for a into the right-hand side of (10), we get
 
                                               (10 + a3)/9a  =   (10 + 23)/9(2)   =   18/18  =   1   =   12   =   c2.
                                                                                                                
          Thus, both (9) and (10) are satisfied by a = 2, c = 1.    This gives us
                                             _        _              _         __                  _                                                      _
                                        a√b + c√d    =   2√1 + 1√ -3    =   2 + i√3     as a cube root of   - 10 + 9i√3.  
 
          Check:    (2 + i√3)3    =    (2 + i√3)( 2 + i√3)( 2 + i√3)
                                               =    (4 + 2i√3 + 2i√3 – 3)(2 + i√3)
                                               =    (1 + 4i√3)(2 + i√3)
                                               =     2 + i√3 + 8i√3 – 12   =  - 10 + 9i√3.
                                                                _
     3.  Find a cube root of 458 + 435√3.      Using the Cube Root Theorem above, we have   
                              _                _             _            _             _
          458 + 435√3   =  458√1 + 435√3  =  R√b   +   S√d,   so   R  =  458,  b = 1,  S  =  435,  and  d  =  3.    
 
          Using these values to find a and c, we get
 
                                                                              a2  =   (S – c3d) / 3bc                
                                                                                    =    (435 – 3c3) / 3c.                                                              (11)
   
           and                                                            c2   =   (R – a3b) / 3ad
                                                                                    =    (458 - a3) / 9a                                                                 (12)
                                                                                                                                                                                                                                         
 
          To find a2, we shall substitute values of c (starting with 1) into (11) until we get a perfect square.
 
                                                                   c                    a2  =  (435 – 3c3)/ 3c
                                                              ----------------------------------------------------------------
                                                                   1             432/3   =   144 (a perfect square)
 
            So, when c = 1 a2 = 144, and a = 12.  Thus, a = 12 and c = 1.   Now, equation (12) must be satisfied
            also, so substituting our value for a into the right-hand side of (12), we get
 
                      (458 – a3)/9a  =  (458 – 123)/9(12)  =  (458 – 1728)/108  =  - 1270/108   ≠  12  =  c2.
                                                                                                                
            Thus, (12) is not satisfied by a = 12, c = 1.   We must continue substituting values of c into (11).
 
                                                                      c                     a2 = (435 – 3c3)/ 3c
                                                               --------------------------------------------------------------------
                                                                      1              432/3  =   144 (a perfect square)
                                                                      2              411/6   =   68.5
                                                                      3              354/9   =   39.3
                                                                      4              243/12 =   20.25
                                                                      5              60/15   =   4 (another perfect square)
 
          So, when c = 5 a2 = 4, and a = 2. Thus, a = 2 and c = 5.     Now, equation (12) must be satisfied also, so
          substituting our value for a into the right-hand side of (12), we get
 
                                                (458 – a3)/9a  =  (458 – 23)/9(2)  =  450 /18  =  25  =  52  =  c2.
 
          Thus, both (11) and (12) are satisfied by a = 2, c = 5.   This gives us
                                     _             _             _        _                    _                                                       _
                                 a√b   +   c√d   =   2√1 + 5√3   =   2 + 5√3    as a cube root of   458 + 435√3. 
 
          Check:    (2 + 5√3)3   =   (2 + 5√3)( 2 + 5√3)( 2 + 5√3)
                                               =   (4 + 20√3 + 75)(2 + 5√3)
                                               =   (79 + 20√3)(2 + 5√3)
                                               =  158 + 395√3 + 40√3 + 300
                                               =  458 + 435√3.
                                                          _            _
     4.   Find a cube root of  - 338√5  –  9i√7.      Using the Cube Root Theorem above, we have   
                       _           _                     _         __            _             _
           - 338√5  –  9i√7    =   - 338√5  – 9√ -7  =   R√b   +   S√d,   so  R  = - 338, b = 5,  S = - 9, and d = - 7.    
 
           Using these values to find a and c, we get
 
                                                                       a2  =   (S – c3d) / 3bc                
                                                                             =   ( - 9 + 7c3) / 15c.                                                                     (13)
   
            and                                                    c2  =   (R – a3b) / 3ad
                                                                            =   (- 338 – 5a3) / (-21a)
                                                                            =    (338  +  5a3) / 21a                                                                  (14)
                                                                                   
          To find a2, we shall substitute values of c (starting with 1) into (13) until we get a perfect square.
 
                                                              c                    a2 = ( - 9 + 7c3) / 15c
                                                        ----------------------------------------------------------------------
                                                              1                         - 2/15
                                                              2                         47/30   =  1.57
                                                              3                       180/45   =   4 (a perfect square)
 
          So, when c = 3 a2 = 4, and a = 2. Thus, a = 2 and c = 3.     Now, equation (14) must be satisfied also,
          so substituting our value for a into the right-hand side of (14), we get
 
                                        (338 + 5a3) / 21a   =   (338 + 40)/42    =    378/42   =   9   =   c2.
 
          Thus, both (13) and (14) are satisfied by a = 2, c = 3.
                                        _            _             _         __             _         _                                             _          _
          This gives us   a√b   +   c√d   =   2√5 + 3√ -7   =   2√5 + 3i√7   as a cube root of - 338√5 – 9i√7. 
 
          Check:   Using the procedure we developed at the beginning of this article for cubing a radical,
                         we have
                                _         _                  _         ___
                          (2√5 + 3i√7)3   =    [2√5 + 3√(-7) ]3,   where a = 2, b = 5, c = 3, and d = - 7,
 
                                                    =    2(                               )√5     +    3(                               )√(-7)
 
                                                    =    2[  22(5)          32(-7)  ]√5    +    3[  22(5)         32(-7)  ]√(-7)
 
                                                    =    2[   20              - 63     ]√5    +    3[   20             - 63    ]√(-7)
 
                                                    =    2[   20    +     3(- 63)   ]√5    +    3[ 3(20)   +    (- 63)   ]√(-7)
 
                                                    =    2[   20    +     (- 189)   ]√5    +    3[    60     +    (- 63)   ]√(-7)
 
                                                    =    2[ -169 ]√5    +    3[ -3 ]√(-7)   =   - 338√5   –  9i√7.
        
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