Find solution set of $\dfrac{8^x+27^x}{12^x+18^x}=\dfrac{14}{12}$ What I've done is factoring it.
$$\dfrac{2^{3x}+3^{3x}}{2^{2x}\cdot 3^{x}+3^{2x}\cdot{2^{x}}}=\dfrac{7}{2\cdot 3}$$
This looks like it can be factored more but it doesn't work from my attempts.

$12(8^x+27^x)=14(12^x+18^x)$
$\Rightarrow $ $6(8^x+27^x)=7(12^x+18^x)$
Divide by$12^x$
$\Rightarrow $$ 6((\frac{2}{3}) ^x+(\frac{3}{2})^{2x}) =7(1+(\frac{3}{2})^{x})$
$\Rightarrow $ $(\frac{2}{3}) ^x+(\frac{3}{2})^{2x}-\frac{7}{6}-(\frac{7}{6})(\frac{3}{2}) ^x=0$
We put :$t=(\frac{3}{2}) ^x$
$\Rightarrow $ $\frac{1}{t} +t^2 - \frac{7}{6}t-\frac{7}{6}=0$
Multiply by $t$
$\Rightarrow $ $ 1+t^3 - \frac{7}{6}t^2-\frac{7}{6} t=0$
We can see $-1$ is a solution of the equation
So:after division by $t+1$ we see
$ t^2 - \frac{13}{6}x+1=0$
$\triangle =(\frac{13}{6})^2 - 4=(\frac{5}{6}) ^2 $
So :
$t_1=\frac{\frac{13}{6}+\frac{5}{6}}{2}=\frac{3}{2} =(\frac{3}{2})^{x_1} $
$\Rightarrow $ $x_1=1$
And
$t_2=\frac{\frac{13}{6}-\frac{5}{6}}{2}=\frac{2}{3}=(\frac{3}{2})^{x_2} $
$\Rightarrow $ $x_2=-1$
Finally
$x=-1$ or $ x= 1$
