Substituting z=abcxy in first three equations,
we have,
ax2−bcx+abc=0
by2−cay+abc=0
x2y2−abxy+abc2=0
From the last two equations we have
y2a2bc(c2−bx)=−yabc(x2−bc)−1ax(b2−cx)
Hence, a3bcx(c2−bx)(b2−cx)=a2b2c2(x2−bc)2
i.e., bcx4−abcx3+x2(ab3+ac3−2b2c2)−ab2c2x+b2c2=0
It remains to eliminate x between this equation and ax3−bcx+abc=0
Multiply the first equation by 'a' and the equation ax2−bcx+abc=0 by bc(bc−a2)x and subtract, then
(a3b3+a3c3+b3c3−4a2b2c2)x2−ab3c3x+a2b3c3=0
Multiply ax2−bcx+abc=0 by
ab2c2 and subtract from this last equation;-
(a3b3+b3c3+c3a3−5a2b2c2)x2=0
∴a3b3+b3c3+c3a3=5a2b2c2