1. The problem is to solve the quartic equation $$y^4 + p y^2 + q y + r = 0$$ by reducing it to a cubic equation in $s$. 2. We start by rewriting the quartic as $$\left(y^2 + \frac{p}{2}\right)^2 = -q y - r + \frac{p^2}{4}$$ and then introduce $s$ to complete the square: $$\left(y^2 + \frac{p}{2} + s\right)^2 = 2 s y^2 - q y + p s + \frac{p^2}{4} + s^2 - r$$ 3. To eliminate the $y$ term, we impose the condition that the discriminant of the quadratic in $y$ is zero: $$q^2 - 4 (2 s) \left(p s + \frac{p^2}{4} + s^2 - r\right) = 0$$ which simplifies to $$8 s^3 + 8 p s^2 + \left(\frac{p^2}{4} - r\right) s - q^2 = 0$$ 4. Dividing by 8, the cubic equation in $s$ is: $$s^3 + p s^2 + \left(\frac{p^2}{4} - r\right) s - q^2 = 0$$ 5. The roots $s_1, s_2, s_3$ of this cubic can be found using the cubic formula involving cube roots and square roots of expressions in $p, q, r$: $$s_1 = -\frac{p}{3} + \sqrt[3]{A + \sqrt{A^2 + B^3}} + \sqrt[3]{A - \sqrt{A^2 + B^3}}$$ where $$A = \frac{p^3}{24} - \frac{p r}{6} - \frac{p^3}{27} + \frac{q^2}{16}, \quad B = \frac{r^2}{12} - \frac{q^2}{9}$$ $$s_2 = -\frac{p}{3} - \sqrt[3]{C + \sqrt{C^2 - D^3}} + \sqrt[3]{C - \sqrt{C^2 - D^3}}$$ where $$C = \frac{p^3}{216} - \frac{p r}{6} + \frac{q^2}{16}, \quad D = \frac{p^2}{36} + r$$ 6. Once $s$ is found, substitute back to solve for $y$ by solving the quadratic: $$\left(y^2 + \frac{p}{2} + s\right)^2 = 2 s y^2 - q y + p s + \frac{p^2}{4} + s^2 - r$$ which reduces to two quadratics in $y$. This method transforms the quartic into a resolvent cubic, whose roots help factor the quartic into quadratics. Final answer: The quartic equation reduces to solving the cubic in $s$ given by $$s^3 + p s^2 + \left(\frac{p^2}{4} - r\right) s - q^2 = 0$$ whose roots $s_1, s_2, s_3$ are given by the formulas above.