This question is an attempt to resuscitate this question, which was recently deleted by its author for reasons unknown to Yours Truly.
The question runs as follows: given the polynomial
$X^3 + 3X^2 + cX + 3 \in \Bbb F_5[X], \tag 1$
find all
$c \in \Bbb F_5 \tag 2$
such that $F_5[X]/(X^3 + 3X^2 + cX + 3)$ is a field.
Various approaches to this problem have been made in the linked citing; my own solution is presented below in my answer.
Note Added in Edit, Monday 15 June 2020 6:53 PM PST: In light of some of the comments, I would like to address the issue of differences/similarities between this question and the one linked in the above. End of Note.
Since
$X^3 + 3X^2 + 3cX + 3 \in \Bbb F_5 \tag 1$
is a cubic polynomial, it is reducible if and only if it has a linear factor, i.e., if and only if it is possessed of a root in $\Bbb F_5$. We may evaluate this polynomial on the elements of $\Bbb F_5$ to determine the values of $c$ for which a zero exists. Setting
$p(X, c) = X^3 + 3X^2 + 3cX + 3, \tag 2$
we have
$p(0, c) = 3, \tag 3$
$p(1, c) = 3c + 2, \tag 4$
$p(2, c) = c + 3, \tag 5$
$p(3, c) = 4c + 3, \tag 6$
$p(4, c) = 2c; \tag 7$
scrutiny of these equations reveals that $p(X, c)$ has a root precisely when
$c = 0, 1, 2, 3; \tag 8$
thus $p(X, c)$ is irreducible for
$c = 4. \tag 9$
For this value of $c$, the ideal $(p(X, c))$ is maximal and hence $\Bbb F_5[X]/(p(X, c))$ is a field.