<s>[INST] $\frac{'ab'}{'ba'}=\frac{'bc'}{'cb'}$ Question: 

We write $'pq'$ to denote a two digit integer with tens digit $p$ and units digit $q$. For which values of $a,b$ aand $c$ are the two fractions $\frac{'ab'}{'ba'}$and$\frac{'bc'}{'cb'}$  equal and different from $1$?

My attempt:

So, write this using "normal algebra";$$\frac{10a+b}{10b+a}=\frac{10b+c}{10c+b}$$Multiply by the two denominators:$$(10a+b)(10c+b)=(10b+c)(10b+a)$$Multiply out:$$100ac+10ab+10bc+b^2=100b^2+10ab+10bc+ac$$ And we can subtract $(10ab + 10bc) $ from each side. We are left with:$$100ac+b^2=100b^2+ac$$ Which, by rearranging, we get $$99ac=99b^2$$which is equivalent to (if we divide by $99$) $$ac=b^2$$
  $$[a,b,c]\neq 0$$ Since they are all parts of the tens digit, and if they are $0$, then they will not be a two digit number, which is vital. $a \neq b$, since if they were $=$, the fraction would $= 1$, and neither can this relationship occur between $b \neq c$

This was all written down on paper by me. Can anybody help me finish this off? 
 [/INST] Looks good so far. Now you just have to write down all the ways that $b^2 = ac$ for $1 \leq a, b, c \leq 9$, $a \neq b$, $b \neq c$ : $$\begin{align*} 2^2 &= 1 \times 4 \\ 3^2 &= 1 \times 9  \\ 4^2 &= 2 \times 8 \\ 6^2 &= 4 \times 9 \end{align*} $$
Each of these gives two solutions; e.g. $2^2 = 1 \times 4$ gives $(a, b, c) = (1, 2, 4)$ or $(a, b, c) = (4, 2, 1)$. The criterion $b^2 = ac$ states that $a, b, c$ stand in a geometric progression; if the common ratio of this progression is $r$, then $'bc' = r \times {'ab'}$ and $'cb' = r \times {'ba'}$, so this should make intuitive sense.
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