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Triple cube sum (Posted on 2019-06-28) Difficulty: 3 of 5
a+b+c=6
(a-b)/c+(b-c)/a+(c-a)/b=(1+a/b)(1+b/c)(1+c/a)
a2/bc+b2/ca+c2/ab+ab/c2+bc/a2+ca/b2=-2
a3+b3+c3=?

No Solution Yet Submitted by Danish Ahmed Khan    
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computer solution | Comment 1 of 3
I hate doing a lot of algebra (although this reminds me of Heron's formula.)
anyway, the answer is  6^3 = 216

ia ib ic       -6           6           6   216.000000   

ia ib ic        6          -6           6   216.000000  

ia ib ic        6           6          -6   216.000000  



program abc

        implicit none

        integer ia,ib,ic

        real a,b,c,lh1,rh1,lh2,cube


           do ia=-1000,1000

                do ib=-1000,1000

                a=ia

                b=ib

                c=6-(ia+ib)

                ic=c

        lh2=a**2/(b*c)+b**2/(c*a)+c**2/(a*b)+a*b/c**2+b*c/a**2+c*a/b**2

                if (lh2.ne.-2.)go to 1

                lh1=(a-b)/c+(b-c)/a+(c-a)/b

                rh1=(1+a/b)*(1+b/c)*(1+c/a)

                if(lh1.ne.rh1)go to 1

c               print*,lh1,lh2,rh1,

        cube=ia**3+ib**3+ic**3

        print*,'ia ib ic ',ia,ib,ic, cube


1               enddo

2          enddo 

        end



Edited on June 28, 2019, 7:31 pm
  Posted by Steven Lord on 2019-06-28 19:30:31

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