Example 18. Solve the equation

$$
3^{2 x^{2}-6 x+3}+6^{x^{2}-3 x+1}=2^{2 x^{2}-6 x+3}
$$
Solution. Equation (17) is equivalent to the equation

$$
27 \cdot 3^{2\left(x^{2}-3 x\right)}+6 \cdot 3^{x^{2}-3 x} 2^{x^{2}-3 x}-8 \cdot 2^{2}\left(x^{2}-3 x\right)=0
$$

the solution of which reduces to solving the homogeneous equation

$$
27 f^{2}(x)+6 f(x) g(x)-8 g^{2}(x)=0
$$

where $f(x)=3^{x^{2}-3 x}, g(x)=2^{x^{2}-3 x}$.
Since $2^{x^{2}-3 x}>0$, the equation (18) can be divided by $2^{2\left(x^{2}-3 x\right)}$. Then we obtain the equation

$$
27(3 / 2)^{2\left(x^{2}-3 x\right)}+6(3 / 2)^{x^{2}-3 x}-8=0
$$

which is equivalent to equation (17).

Let $t=(3 / 2)^{x^{2}-3 x}$; then from the equation $27 t^{2}+6 t-8=0$ we find $t_{1}=-2 / 3, t_{2}=4 / 9$.

Thus, equation (17) is equivalent to the equation

$$
(3 / 2)^{x^{2}-3 x}=(3 / 2)^{-2}
$$

i.e.

$$
x^{2}-3 x+2=0
$$

From this, we find $x_{1}=1, x_{2}=2$-the roots of equation (17).