How to use Pumping Lemma for $L = \{www | w∈\{0,1\}^*\}$
Let's take the alphabets as a and b in-order to avoid confusion.

Suppose that A1 is a regular language. Let p be the “pumping length”
of the Pumping Lemma. Consider the string $s = a^pba^pba^pb$. Note that $s ∈ A1$
since $s = (a^pb)^3$, and $|s| = 3(p + 1) ≥ p$, so the Pumping Lemma will hold.

Thus, we can split the string $s$ into 3 parts $s = xyz$ satisfying the conditions.

i. $xy^iz ∈ A1$ for each $i ≥ 0$

ii. $|y| &gt; 0$,

iii.$ |xy| ≤ p.$

Since the first $p$ symbols of $s$ are all $a’s$, the third condition implies that $x$ and $y$ consist only of $a’s$. So $z$ will be the rest of the first set of $a’s$, followed by $ba^pba^pb$.

The second condition states that $|y| &gt; 0$, so $y$ has at least one $a$. More precisely, we can then say that

$x = a^j$ for some $j ≥ 0$,

$y = a^k$ for some $k ≥ 1$,

$z = a^mba^pba^pb$ for some $m ≥ 0$.

Since $a^pba^pba^pb = s = xyz = a^ja^ka^mba^pba^pb = a^{j+k+m} ba^pba^pb$, we must have that $j + k + m = p$. The first condition implies that $xy^2z ∈ A1$, but

$xy^2z = a^ja^ka^ka^mba^pba^pb$ 

$ = a^{p+k}ba^pba^pb$

since $j + k + m = p$. Hence, $xy^2z$ not an element of $A1$ because $k ≥ 1$, and we get a contradiction. Therefore, $A1$ is a nonregular language.