Let's determine all integer values of $x$ for which

$$
\frac{8^{x}+27^{x}}{12^{x}+18^{x}}=\frac{7}{6}
$$
I. solution. If $\mathrm{x}=1$, the value of the left side is exactly $\frac{7}{6}$.

$$
\text{For } x \geq 2
$$

$$
\frac{27^{x}+8^{x}}{18^{x}+12^{x}}>\frac{27^{x}}{18^{x}}-1
$$

therefore, the left side of (1) is at least $\left(\frac{3}{2}\right)^{2}-1=1+\frac{1}{4}$, which is greater than the $\left(1+\frac{1}{6}\right)$ on the right side. If $x=0$, the value of the left side is 1, and if $x=-y(18 \cdot 27)^{x}+(12 \cdot 27)^{x}-(18 \cdot 18)^{x}-(12 \cdot 18)^{x}
$$

we get the inequality, which is indeed true, since on the right side

$$
(12 \cdot 27)^{x}=(18 \cdot 18)^{x}
$$

II. solution. Let the value of $2^{x}$ and $3^{x}$ be denoted by $a$ and $b$, respectively, then the value of the fraction on the left side of (1) is

$$
\frac{a^{3}+b^{3}}{a^{2} b+a b^{2}}=\frac{(a+b)\left(a^{2}-a b+b^{2}\right)}{a b(a+b)}=\frac{a}{b}-1+\frac{b}{a}
$$

thus, (1) gives the equation

$$
c+\frac{1}{c}=\frac{13}{6}
$$

for the unknown $c=\frac{a}{b}$. From this, we get the quadratic equation $6\left(c^{2}+1\right)=13 c$, whose roots are $c_{1}=\frac{3}{2}$ and $c_{2}=\frac{2}{3}$, so the value of the

$$
c=\frac{a}{b}=\left(\frac{2}{3}\right)^{x}
$$

power is either $\frac{2}{3}$ or its reciprocal, hence the value of $x$ is either 1 or -1.