Being simultaneously an arithmetic progression and a geometric one

Question

Let $a,b,c$ be three consecutive terms of an arithmetic progression and a geometric one($a \neq 0$).Find the value of: $$\frac{a^3b+2b^2c^2-c^4}{a^2b^2+b^4-3c^4}$$

My attempts:Letting $a=b=c$ shows us that the answer is $-2$.Writing $a,b,c$ using the geometric progression $a,aq,aq^2$ we will get into an expression by $q$ which I don't know how to finish from here knowing that $a,b,c$ form an arithmetic progression.

Answer 1

If $a,b,c$ are in arithmetic progression, then $b-a=c-b$ or $2b=a+c.$

If $a,b,c$ are in geometric expression, then $\frac{b}{a}=\frac{c}{b},$ or $b^2=ac.$

This means $(a+c)^2=4ac.$ Show that if $(a+c)^2=4ac,$ then $a=c.$

Answer 2

We have given $$a,b=aq,c=aq^2$$ and $$a,b=a+d,c=a+2d$$ so we get $$aq=a+d$$ and $$aq^2=a+2d$$ eliminating $d$ we get $$aq^2=2aq-a$$ since $$a\neq 0$$ we get for $q$ the equation $$(q-1)^2=0$$ Can you finish?