Answer
$\color{blue}{p_X(k) = (k^3-(k-1)^3)/216,\ k=1,2,3,4,5,6.}$
Work Step by Step
The sample space for the faces showing in the toss of three fair dice is $\{(i,j,k)\ \vert\ i,j,k \in \{1,2,3,4,5,6\}\}$ as each die can come up either a $1,2,3,4,5$, or $6.$
The sample space has $6^3 = 216$ elements.
If the largest of the three faces that appear in the throw of three dice is $k$, then $k \in \{1,2,3,4,5,6\}$.
The total number of ways in which the largest of the three faces that appear in the throw of three dice is $k$, is equal to the number of ways in which the face that appears on each of the three dice is at most $k$ less the number of ways in which the face that appears on each of the three dice is at most $k-1$. The former event can occur in $k^3$ ways, as each of three dice can come up in $k$ ways (one of $1,2,\ldots,k$), while the latter event can occur in $(k-1)^3$ ways, as each of the three dice can up in $k-1$ ways (one of $1,2,\ldots, k-1$).
Therefore, the total number of ways in which the largest of the three faces that appear in the throw of three dice is $k$, is $k^3-(k-1)^3,\ k=1,2,3,4,5,6$.
Thus, if we let $X$ be the random variable for the largest of the three faces that appear in the throw of three dice, $P(X=k) = (k^3-(k-1)^3)/216$ since the total number of successful events is $k^3-(k-1)^3$, while the total sample space has $216$ elements. Consequently, the pdf of $X$ is
$\begin{align*}
p_X(k) &= P(X=k),\ k=1,2,3,4,5,6 \\
\color{blue}{p_X(k)} &\color{blue}{= (k^3-(k-1)^3)/216,\ k=1,2,3,4,5,6}.
\end{align*}$