$a^{\frac{3}{4}}$ is to be factored into a product of 4 factors, so that the exponent of each factor is $\frac{1}{6}$ greater than the exponent of the preceding factor.
Let the first factor be $a^{x}$, then

$$
a^{\frac{3}{4}}=a^{x} \cdot a^{x+\frac{1}{6}} \cdot a^{x+\frac{2}{6}} \cdot a^{x+\frac{3}{6}}
$$

which means

$$
\frac{3}{4}=4 x+1
$$

from which

$$
x=-\frac{1}{16}
$$

so

$$
a^{-\frac{1}{16}}, a^{\frac{6}{18}}, a^{\frac{13}{48}}, \text { and } a^{\frac{7}{16}}
$$

are the sought factors.

(Paunz Rezsó, Pécs.)