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?
 A: 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.
A: $$\frac{a^3 + b^3}{a^2b + ab^2} = \frac{7}{6} \implies 7 a^2b + 7ab^2 = 6 a^3 + 6b^3 \implies 7ab(a + b) = 6(a + b)(a^2 -a b + b^2)$$ $$ \implies (a + b)(6a^2 -13ab + 6b^2) =0 $$
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.
A: 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 $x=\pm1$.
A: You properly wrote $$\frac{a^3+b^3}{a^2b+ab^2} = \frac{7}{6}$$ Tak into account the homegeneity and let $b=k a$ to get
$$k+\frac 1k-1=\frac{7}{6}\implies k=\frac 23 \qquad \text{and} \qquad k=\frac 32$$ and then the slutions.
