Find all real numbers $x$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac76$

Let us put $2^x=a,3^x=b$ to remove the indices to improve clarity

So, $8^x=(2^3)^x=(2^x)^3=a^3$ and similarly, $27^x=b^3$

$12^x=(2^2\cdot3)^x=(2^x)^2\cdot3^x=a^2b$ and similarly, $18^x=ab^2$

So, the problem reduces to $$\frac{a^3+b^3}{ab(a+b)}=\frac76$$

$\displaystyle\implies 6(a^2-ab+b^2)=7ab$ as $a+b\ne0$

$\displaystyle\implies 6\left(\frac ab\right)^2-13\cdot\frac ab+6=0$

$\displaystyle\implies \frac ab=\frac32$ or $\dfrac23$

So, $\displaystyle\left(\frac23\right)^x=\frac32$ or $\dfrac23$

If $\displaystyle\left(\frac23\right)^x=\frac32\implies\left(\frac23\right)^x=\left(\frac23\right)^{-1}\iff\left(\frac23\right)^{x+1}=1$

Similarly, if $\displaystyle\left(\frac23\right)^x=\frac23, \left(\frac23\right)^{x-1}=1 $

Now if $\displaystyle u^m=1,$

either $\displaystyle m=0,u\ne0; $

or $\displaystyle u=1$

or $\displaystyle u=-1,m$ is even

$$ \begin{array}{rcl} 6(8^x + 27^x) &=& 7(12^x + 18^x) \\ 6(2^{3x} + 3^{3x}) &=& 7(3^x2^{2x} + 3^{2x}2^x) \\ \end{array} $$

Substitute $a\!=\!2^x$ and $b\!=\!3^x$ for simplicity:

$$ \begin{array}{rcl} 6(a^3 + b^3) &=& 7(a^2b + ab^2) \\ 6(a+b)(a^2 - ab + b^2) &=& 7ab(a+b) \\ 6(a^2 - ab + b^2) &=& 7ab \\ 6a^2 -13ab + 6b^2 &=& 0 \\ a^2 -\frac{13}{6}ab + b^2 &=& 0 \end{array} $$

The left hand side can be factorized as $\left(a-\dfrac{3}{2}b\right)\left(a-\dfrac{2}{3}b\right)$.

$$ \left(a-\dfrac{3}{2}b\right)\left(a-\dfrac{2}{3}b\right)=0 \\ \begin{array}{rclcrcl} 2a &=& 3b &\text{or}& 3a &=& 2b \\ 2.2^x &=& 3.3^x &\qquad\text{or}\qquad& 3.2^x &=& 2.3^x \\ x &=& -1 &\qquad\text{or}\qquad& x &=& +1 \end{array} $$

Therefore, $x$ can be either $-1$ or $+1$.

$$ \boxed{x = \mp 1} $$