Express each of X, Y and Z in terms of a,b,c,p,q and r so that the equation given below becomes an identity.

(a³+b³+c³ - 3abc)(p³+q³+r³ - 3pqr) = X³+Y³+Z³ - 3XYZ

Note: Disregard any permutations. For example, if (X, Y, Z) = (α, β, γ), then (X, Y, Z) = (β, γ, α) is invalid.

Let {a,b,c} = {(k-m)^3+(k-n)^3,  (m-k)^3+(m-n)^3,  (n-k)^3+(n-m)^3}            
           
Substituting in (a^3+b^3+c^3 - 3abc) gives:           
           
((k-m)^3+(k-n)^3)^3+((m-k)^3+(m-n)^3)^3+((n-k)^3+(n-m)^3)^3-3 ((k-m)^3+(k-n)^3) ((m-k)^3+(m-n)^3) ((n-k)^3+(n-m)^3) = 0 [1] (Réalis)

     
Given that this is true for any and all k,m,n, there is no one general solution for X, Y and Z in terms of a,b,c,p,q and r.    

       
However, since:           
 (((2k-2m)^3+(2k-2n)^3)^3)/((k-m)^3+(k-n)^3)^3) = 512,

        
and in like manner:           
((3k-3m)^3+(3k-3n)^3)^3/((k-m)^3+(k-n)^3)^3 = 19683,  

it follows that:         
           
a = {(k-m)^3+(k-n)^3,  b = (m-k)^3+(m-n)^3, c = (n-k)^3+(n-m)^3}            
p = 512a , q = 512b, r = 512c            
X = 19683a , Y = 19683b, Z = 19683c     

       
is a solution. 

e.g: {k,m,n} = {-2,0,3}; {a,b,c} = -133,-19, 152; a^3+b^3+c^3 = 1152312; 3abc = 1152312         
{p,q,r} = {-68096, -9728, 77824}; p^3+q^3+r^3 = 154660698587136, 3pqr = 154660698587136         
{X,Y,Z} = {-2617839, -373977, 2991816}; X^3+Y^3+Z^3 = 8787067489120339944, 3XYZ= 8787067489120339944

         
(0*0) =0 

 

Edited on November 3, 2013, 2:19 am

Posted by broll on 2013-11-03 01:38:50