Manipulating an algebraic equation This came in a competition, I went back home and tried it but was unsuccessful. 
If $$\frac{a^3+3ab^2}{3a^2b+b^3} = \frac{x^3+3xy^2}{3x^2y+y^3}$$ then -
$$1) ~ax =by
~~2)~ xy = ab
~~3)~ ay = bx
~~4) ~ax = b$$
It is clear that if the following given condition holds for the variables then there would be a case in which $a  = x$ and $b = y$. Hence then options should also satisfy this and hence we can eliminate options a and d. But I was not able to solve the question further. I tried adding and subtracting 1 from both sides but found nothing helpful. 
Moreover we are required to show if the correct option is sufficient to show that the first condition is true for the given variables. 
 A: Apply componendo-dividendo on this :
$$\frac{a^3+3ab^2}{3a^2b+b^3} = \frac{x^3+3xy^2}{3x^2y+y^3} $$
To get :
$$\frac{(a+b)^3}{(a-b)^3}=\frac{(x+y)^3}{(x-y)^3}$$
$$\implies \frac{(a+b)}{(a-b)}=\frac{(x+y)}{(x-y)}$$
Now apply reverse componendo-dividendo to get $$\frac{a}{b}=\frac xy$$
That is :
$$\color{blue} {ay=bx}$$

Componendo-dividendo is : 
If you have the ratio $$\frac pq =\frac rs$$ Then you can add $1$, and subtract $1$ (In both, LHS and RHS) and take the ratio of two equations thus formed to get $$\frac{p+q}{p-q}=\frac{r+s}{r-s}$$
A: $$
\frac{a(a^2+3b^2)}{b(3a^2+b^2)}=\frac{x(x^2+3y^2)}{y(3x^2+y^2)}
$$
Define
$$\alpha=a/b,\;\;\;\chi=x/y$$
$$
\frac{\alpha(\alpha^2+3)}{3\alpha^2+1}=\chi\frac{\chi^2+3}{3\chi^2+1}
$$
The function
$$
f(x)=x\frac{x^2+3}{3x^2+1}
$$
is increasing (a.e.), that is if $x\neq y$ then $f(x)\neq f(y)$. Since the original equation you gave reads $f(\alpha)=f(\chi)$ we can conclude that $\alpha=\chi$ or in terms of $a,b,x,y$:
$$
\frac{a}{b}=\frac{x}{y}
$$
cross multiplying
$$
ay=bx
$$
A: $$0=\frac{a^3+3ab^2}{3a^2b+b^3} -\frac{x^3+3xy^2}{3x^2y+y^3}=$$
$$=\frac{(ay-bx)(3a^2x^2+3b^2y^2+a^2y^2+b^2x^2-8abxy)}{(3a^2b+b^3)(3x^2y+y^3)}=$$
$$\frac{(ay-bx)(3(ax-by)^2+(ay-bx)^2)}{(3a^2b+b^3)(3x^2y+y^3)},$$
which says that $ay=bx$ or
$$(ax-by)^2+(ay-bx)^2=0.$$
In the last case we  obtain $ay=bx$ again.
Thus, the answer is $ay=bx$.
A: Rewrite $$\frac{a^3+3ab^2}{3a^2b+b^3} = \frac{x^3+3xy^2}{3x^2y+y^3}\tag 1$$ as 
$$\frac{ ({a \over b})^3+3{a \over b}}{3({a\over b})^2+1} = \frac{({x \over y})^3+3{x\over y}}{3({x\over y})^2+1} \tag 2$$ and set $t={a\over b},\; s={x\over y}.$ Then
$$0=\frac{t^3+3t}{3t^2+1}-\frac{s^3+3s}{3s^2+1}=\frac{(t-s)\left[(t-s)^2+3(ts-1)^2\right]}{(3t^2+1)(3s^2+1)}.$$ Consequently $t=s,$ which is ${a \over b} = {x \over y},$ or $$ax=by \tag 3.$$
Sufficiency
Assume $ax=by.$ If $b\neq 0 \neq y,$ then clearly $(2)$ holds and $(1)$ follows.
 However, $ax=by$ is true if $b=0=y$ as well, but  $(1)$ is not defined.
