Let \(k\) be a field and let \[ f(x)=x^{4}+a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0} \in k[x]. \] The resolvent cubic of \(f\) is \[ R_{4}(t) \coloneqq t^{3}-a_{2} t^{2}+\left(a_{1} a_{3}-4 a_{0}\right) t+4 a_{0} a_{2}-a_{1}^{2}-a_{0} a_{3}^{2} \in k[t]. \]
Let \(f\in k[x]\) be as in the definition, and let \(r_1, r_2, r_3, r_4\) be the roots of \(f\), with multiplicity, in a splitting field of \(f\) over \(k\). Then \[ R_4(t) = \left(t-\left(r_{1} r_{2}+r_{3} r_{4}\right)\right)\left(t-\left(r_{1} r_{3}+r_{2} r_{4}\right)\right)\left(t-\left(r_{1} r_{4}+r_{2} r_{3}\right)\right). \]
For \(f(x) = x^4 + cx + d\), that is, \(a_3 = a_2 = 0\), \(a_1 = c\), and \(a_0 = d\), the resolvent cubic is \(R_4(t) = t^3 - 4dt - c^2\).