STEP II 1992 question 7 solution

\text{Suppose } a=b, \text{ then the equation becomes} \\
a^2c = r,\quad a^2+2ac=q,\quad 2a+c=0.\\
4q^3+27r^2 = 4(a^2+2ac)^3 + 27(a^2c)^2\\
=4(a^6 + 6a^5c + 12a^4c^2 + 8a^3c^3) + 27a^4c^2\\
= 4a^6 + 24a^5c + 48a^4c^2 + 32a^3c^3 + 27a^4c^2.\\
\text{But } c = 2a \Rightarrow 4q^3 + 27r^2 = 4a^6 - 24a^6 + 192a^6 - 256a^6 + 108a^6\\

= \frac{4(2ac-a^2-c^2)^3}{3^3} + 27\left( \frac{-2a^3 + 6a^2c - 6ac^2 + 2c^3}{27} \right) ^2\\
= \frac{4(-1)^3((a-c)^2)^3}{3^3} + \frac{27(-2)^2((a-c)^3)^2}{27^2}\\