The case is of three masses *A*,*B*,*C* at the vertices of a triangle with sides *a*,*b*,*c*.
It is required to show that the triangle is either collinear or equilateral.
The analysis gets as far as an equation labelled (52a):

2*Ka*^{3}*b*^{3}*c*^{3}/*G*=2(*C*+*B*)*c*^{3}*b*^{3}+*Aa*[*c*^{3}(*a*^{2}+*b*^{2}−*c*^{2})+*b*^{3}(*c*^{2}+*a*^{2}−*b*^{2})]

with equations (52b) and (52c) obtained by cyclic permutation.

*To continue: *The LH sides are all equal; hence so are the RH sides.
From each RH side subtract the cyclically symmetric expression

*Aa*^{3}*b*^{3}+*Aa*^{3}*c*^{3}+*Bb*^{3}*c*^{3}+*Bb*^{3}*a*^{3}+*Cc*^{3}*a*^{3}+*Cc*^{3}*b*^{3}

to get three equal expressions

(*B*+*C*)*b*^{3}*c*^{3}+*Aac*^{3}(*b*^{2}−*c*^{2})+*Aab*^{3}(*c*^{2}−*b*^{2})−*Bb*^{3}*a*^{3}−*Cc*^{3}*a*^{3}=...

Negating and rearranging each expression gives

*Aa*(*b*^{2}−*c*^{2})(*b*^{3}−*c*^{3})+*Bb*^{3}(*a*^{3}−*c*^{3})+*Cc*^{3}(*a*^{3}−*b*^{3})=...

Put *A*_{1}=*Aa*(*b*^{3}−*c*^{3}), *B*_{1}=*Bb*(*c*^{3}−*a*^{3}), *C*_{1}=*Cc*(*a*^{3}−*b*^{3}) to get

*A*_{1}(*b*^{2}−*c*^{2})−*B*_{1}*b*^{2}+*C*_{1}*c*^{2}=*B*_{1}(*c*^{2}−*a*^{2})−*C*_{1}*c*^{2}+*A*_{1}*a*^{2}=*C*_{1}(*a*^{2}−*b*^{2})−*A*_{1}*a*^{2}+*B*_{1}*b*^{2}.

Next put *A*_{2}=*B*_{1}−*C*_{1}, *B*_{2}=*C*_{1}−*A*_{1}, *C*_{2}=*A*_{1}−*B*_{1} to get

*b*^{2}*C*_{2}+*c*^{2}*B*_{2}=*c*^{2}*A*_{2}+*a*^{2}*C*_{2}=*a*^{2}*B*_{2}+*b*^{2}*A*_{2}.

These equations can be solved for *A*_{2},*B*_{2},*C*_{2} up to multiplication by a constant. The result is

*A*_{2}=*ka*^{2}(−*a*^{2}+*b*^{2}+*c*^{2}), *B*_{2}=*kb*^{2}(*a*^{2}−*b*^{2}+*c*^{2}), *C*_{2}=*kc*^{2}(*a*^{2}+*b*^{2}−*c*^{2})

where *k* is arbitrary (we use the fact that −*a*^{2}+*b*^{2}+*c*^{2} etc. are not all 0).

But *A*_{2}+*B*_{2}+*C*_{2}=0, so that

*k*(2*b*^{2}*c*^{2}+2*c*^{2}*a*^{2}+2*a*^{2}*b*^{2}−*a*^{4}−*b*^{4}−*c*^{4})=0.

The polynomial in *a*,*b*,*c* is 16Δ^{2} where Δ is the area of the triangle (Heron’s formula).
Hence either Δ=0, i.e. the triangle is collinear, or *k*=0.
If *k*=0 then *A*_{2}=*B*_{2}=*C*_{2}=0, hence *A*_{1}=*B*_{1}=*C*_{1}, i.e. by definition

*Aa*(*b*^{3}−*c*^{3})=*Bb*(*c*^{3}−*a*^{3})=*Cc*(*a*^{3}−*b*^{3}).

Here the coefficients *Aa*,*Bb*,*Cc* are all positive; hence the terms *b*^{3}−*c*^{3} etc. are either all 0
or all of the same sign. The latter is ruled out since the terms sum to 0.
Hence in the case *k*=0 the triangle is equilateral.