Language of decimal encodings of cubes is not regular
Denote your language by $L$.
Suppose that the pumping length is $p$. Consider
$$
N = (10^{p+1} + 1)^3 = 10^{3p+3} + 3 \cdot 10^{2p+2} + 3 \cdot 10^{p+1} + 1,
$$
whose decimal encoding is
$$
w = 1 0^p 3 0^p 3 0^p 1 \in L.
$$
According to the pumping lemma, we can write $w = xyz$ so that $|xy| \leq p$, $y \neq \epsilon$, and $xz \in L$. Since $|xy| \leq p$, the part $xy$ lies within the prefix $10^{p-1}$. If $y$ contains $1$ then $xz$ has a leading zero, which is impossible (if you allow leading zeroes, consider $L \cap (1+9)(0+9)^*$ instead), so $y$ consists only of zeroes. Hence
$$
xz = 10^q 30^p 30^p 1,
$$
for some $q &lt; p$,
which encodes the number
$$
n = 10^{2p+q+3} + 3 \cdot 10^{2p+2} + 3 \cdot 10^{p+1} + 1.
$$
If $xz \in L$ then $n$ is a cube. Since $n$ ends with $1$, necessarily $n = 10m + 1$ (you can check this directly by considering $0^3,1^3,\dots,9^3$), and so
$$
10^{2p+q+3} + 3 \cdot 10^{2p+2} + 3 \cdot 10^{p+1} + 1 =
1000 m^3 + 300 m^2 + 30m + 1,
$$
which implies that
$$
(10^{p+q+2} + 3 \cdot 10^{p+1} + 3) 10^p =
(100 m^2 + 30 m + 3) m.
$$
Clearly $100m^2 + 30m + 3$ is divisible by neither $2$ nor $5$ (its units digit is $3$), and so $10^p$ must divide $m$. In particular, $m \geq m_0 := 10^p$. However, this is impossible, since $N = (10m_0 + 1)^3$, and $n &lt; N$.