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.
 A: Well you've reached a good point:
$$\dfrac{2^{3x}+3^{3x}}{2^{2x}\cdot 3^{x}+3^{2x}\cdot{2^{x}}}$$
$$=\dfrac{2^{3x}+3^{3x}}{2^{x}\cdot 3^{x}(2^x+3^{x})}$$
We can let $2^x=a$ and $3^x=b$ and get
$$\frac{a^3+b^3}{ab(a+b)}=\frac{a^2-ab+b^2}{ab}=\frac{a}{b}-1+\frac{b}{a}$$
Now let $z=\frac{a}{b}$, we get
$$z-1+\frac{1}{z}=\frac{7}{6} \iff z^2-\frac{13}{6}z+1=0 \implies z=\frac{2}{3},\frac{3}{2} \implies x=\pm 1$$
A: Rearrange the equation 
$$6(8^x+27^x)-7(12^x+18^x)=0$$
and factorize 
$$(2^x+3^x)(3\cdot 3^x - 2\cdot 2^x)(2\cdot 3^x - 3\cdot 2^x)=0
$$ 
which leads to the solutions $x=\pm1$.
A: Let's denote $y=2^x$ and $z=3^x$. Then we have
$$\dfrac{y^3+z^3}{y^2z+z^2y}$$
A: $$\dfrac{2^{3x}+3^{3x}}{2^{2x}\cdot 3^{x}+3^{2x}\cdot{2^{x}}}=\dfrac{7}{2\cdot 3}$$
$$3\cdot 2^{3x+1}+2\cdot 3^{3x+1}=7\cdot 2^{2x}\cdot 3^{x}+7\cdot 3^{2x}\cdot 2^{x}$$
Divide both the sides by $2^{3x+1}$, we get
$$3+3 \left(\frac{3}{2}\right)^{3x}=\frac72 \left(\frac{3}{2}\right)^{x}+\frac72 \left(\frac{3}{2}\right)^{2x}$$
Let $\left(\frac{3}{2}\right)^x=t\quad \forall \ \ t>0$
$$3+3t^3=\frac{7}{2}t+\frac{7}{2}t^2$$
$$6t^3-7t^2-7t+6=0$$
$$(t+1)(3t-2)(2t-3)=0$$
$$t=\frac32 \implies \left(\frac32\right)^x=\frac{3}{2}\iff x=1$$
$$t=\frac 23\implies \left(\frac32\right)^x=\frac{2}{3}\iff x=-1$$
A: $12(8^x+27^x)=14(12^x+18^x)$
$\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$
