Find all $x\in \mathbb{R}$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$ 
Find all $x\in \mathbb{R}$ for which $$\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$$

Letting $a=2^x$ and $b=3^x$ we get $$\frac{a^3+b^3}{a^2b+ab^2} = \frac{7}{6}$$
from the numerator we have that $$a^3+b^3=(a+b)(a^2-ab+b^2)=7$$
since $7$ is a prime we can say that $$a+b=1, a^2-ab+b^2=7.$$
It follows that $$a=1-b$$
from where $$(1-b)^2-(1-b)b+b²=7$$
this quadratic has solutions $b=1, b=0.$
what I now did was consider cases. Firstly $a=b=0$ which has no solutions for $x$.
case $a=b=1$ has the solution $x=0$.
However the actual solutions for this were $x=1, x=-1$ which I don't see how they came up with. What is wrong with my approach?

The problem here is that $a+b$ and $a^2 - ab + b^2$ are not necessarily integers.
@lab bhattacharjee gives a complete answer for the same problem.  In short, observe that both $a^3+b^3$ and $a^2b + ab^2$ are homogeneous of degree $3$. Dividing by $a^3$ yields $$\frac{7}{6}=\frac{a^3+b^3}{a^2b+ab^2}=\frac{1+t^3}{t+t^2}=\frac{1-t+t^2}{t}$$ where $t=\frac{b}{a}=\left( \frac 3 2  \right)^x$. You can proceed from here.
