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?

$$\frac{a^3 + b^3}{a^2b + ab^2} = \frac{7}{6} \implies 7 a^2b + 7ab^2 = 6 a^3 + 6b^3 \implies 7ab(a + b) = 6(a + b)(a^2 -a b + b^2)$$ $$ \implies (a + b)(6a^2 -13ab + 6b^2) =0 $$
Then either $(a + b) = 0$ or $(6a^2 -13ab + 6b^2) = (2a -3b)(3a - 2b) = 0 $
Hence, we have that
$$ a = - b \implies 2^x = - 3^x \implies \left(\frac{2}{3}\right)^x = -1 $$ which is not possible. Hence,
$$ 2a = 3b \implies 2^{x+ 1} =  3^{x +1 } \implies \left(\frac{2}{3}\right)^{x + 1} = \left(\frac{2}{3} \right)^0 \implies x = -1$$
Or, $$ 3a = 2b \implies 2^{x- 1} =  3^{x -1 } \implies \left(\frac{2}{3}\right)^{x + 1} = \left(\frac{2}{3} \right)^0 \implies x = 1 $$
Thus, $x = 1, -1$ are the only soluions.
