Let $x=a+b$, then \begin{align*} x^3 &= 3ab(a+b) + (a^3 + b^3)\\ &= 3abx + (a^3 + b^3)\\ &= 3px + 2q \end{align*} so $p=ab$ and $2q=a^3 + b^3$. Solving for $a$ and $b$: \begin{align*} b &= \frac{p}{a}\\ 2q &= a^3 + \left(\frac{p}{a}\right)^3\\ 2qa^3 &= (a^3)^2 + p^3\\ (a^3)^2 - 2q (a^3) + p^3 &= 0\\ a^3 &= \frac{2q \pm\sqrt{4q^2 - 4p^3}}{2}\\ &= q \pm\sqrt{q^2 - p^3}. \end{align*} $b^3$ has the same solution (because $a$ and $b$ are interchangeable in the original equation), so we can choose $a^3=q+\sqrt{q^2 - p^3}$ and $b^3 = q-\sqrt{q^2 - p^3}$. Then we can add their cube roots together to get $x$: \[x = \sqrt[3]{q+\sqrt{q^2 - p^3}} + \sqrt[3]{q-\sqrt{q^2 - p^3}}.\] When the determinant, $q^2-p^3$, is negative, there will be no solution using this method. In Question 11.4.23, we’ll find a method that will give a solution when the determinant is negative.