Solve the following equation:

$$
8^{x}+27^{x}+64^{x}+125^{x}=24^{x}+30^{x}+40^{x}+60^{x} .
$$
On the left side of the equation, the sum of the cubes of the numbers $a=2^{x}, b=3^{x}, c=4^{x}, d=5^{x}$ stands, while on the right side, $24^{x}=2^{x} \cdot 3^{x} \cdot 4^{x}=a b c, 30^{x}=a b d, 40^{x}=a c d$ and $60^{x}=b c d$. Thus, our equation is of the form

$$
a^{3}+b^{3}+c^{3}+d^{3}=a b c+a b d+a c d+b c d
$$

where $a, b, c, d$ are positive numbers. However, according to the relationship between the arithmetic and geometric means, for all positive $a, b, c, d$,

$$
\frac{a^{3}+b^{3}+c^{3}}{3} \geqq a b c
$$

$$
\frac{a^{3}+b^{3}+d^{3}}{3} \geqq a b d
$$

$$
\frac{a^{3}+c^{3}+d^{3}}{3} \geqq a c d
$$

$$
\frac{b^{3}+c^{3}+d^{3}}{3} \geqq b c d
$$

hold true. Adding these four inequalities, we get that

$$
a^{3}+b^{3}+c^{3}+d^{3} \geqq a b c+a b d+a c d+b c d
$$

equality holds precisely when equality holds in (1), (2), (3), and (4), i.e., when $a=b=c=d$.

The solutions to our equation are thus those $x$ for which $2^{x}=3^{x}=4^{x}=5^{x}$. From $4^{x}=2^{x}$, it is equivalent to the condition $2^{x}=1$, and due to the strict monotonicity of the $2^{x}$ function, this holds uniquely for $x=0$. Indeed, $x=0$ is a solution, as then both sides of the equation are 4.