<s>[INST] 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?
 [/INST] We have that
$$\frac{8^x+27^x}{12^x+18^x}=\frac{(2^x+3^x)(2^{2x}-6^x+3^{2x})}{(6^x)(2^x+3^x)}=\frac{2^{2x}-6^x+3^{2x}}{6^x}=\frac76$$
$$\iff\left(\frac23\right)^x+\left(\frac32\right)^x=\frac{13}6$$
$$\iff y+\frac1y=\frac{13}6 \iff y^2-\frac{13}6y+1=0$$
which leads to $y=\frac23,\frac32$ and $x=\pm1$.
</s>