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):
2Ka3b3c3/G=2(C+B)c3b3+Aa[c3(a2+b2−c2)+b3(c2+a2−b2)]
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
Aa3b3+Aa3c3+Bb3c3+Bb3a3+Cc3a3+Cc3b3
to get three equal expressions
(B+C)b3c3+Aac3(b2−c2)+Aab3(c2−b2)−Bb3a3−Cc3a3=...
Negating and rearranging each expression gives
Aa(b2−c2)(b3−c3)+Bb3(a3−c3)+Cc3(a3−b3)=...
Put A1=Aa(b3−c3), B1=Bb(c3−a3), C1=Cc(a3−b3) to get
A1(b2−c2)−B1b2+C1c2=B1(c2−a2)−C1c2+A1a2=C1(a2−b2)−A1a2+B1b2.
Next put A2=B1−C1, B2=C1−A1, C2=A1−B1 to get
b2C2+c2B2=c2A2+a2C2=a2B2+b2A2.
These equations can be solved for A2,B2,C2 up to multiplication by a constant. The result is
A2=ka2(−a2+b2+c2), B2=kb2(a2−b2+c2), C2=kc2(a2+b2−c2)
where k is arbitrary (we use the fact that −a2+b2+c2 etc. are not all 0).
But A2+B2+C2=0, so that
k(2b2c2+2c2a2+2a2b2−a4−b4−c4)=0.
The polynomial in a,b,c is 16Δ2 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 A2=B2=C2=0, hence A1=B1=C1, i.e. by definition
Aa(b3−c3)=Bb(c3−a3)=Cc(a3−b3).
Here the coefficients Aa,Bb,Cc are all positive; hence the terms b3−c3 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.