Solution to the Depressed Cubic Equation

Theorem 9:  Depressed Cubic Equation

If $x^3 + bx = c$, then:

$$x = \sqrt[3]{\frac{c}{2} + \sqrt{\dfrac{c^2}{4} + \frac{b^3}{27}}} + \sqrt[3]{\frac{c}{2} - \sqrt{\dfrac{c^2}{4} + \frac{b^3}{27}}}$$

Proof:

(1)  $x^3 = -bx + c$

(2)  There exists real numbers $u,v$ such that $x = u + v$

(3)  $(u+v)^3 = u^3 + 3u^2v + 3uv^2 + v^3 = 3uv(u+v) + (u^3 + v^3)$

(4)  Since $b = -3uv$, it follows that $v = \dfrac{-b}{3u}$

(5)  $c = u^3 + v^3 = u^3 + \dfrac{-b^3}{27u^3}$

(6)  It follows that:

$$u^6  - u^3c  + \dfrac{-b^3}{27}= 0$$

(7)  Using the solution for the quadratic equation, we have:

$$u^3 = \frac{c \pm \sqrt{c^2 + \frac{4b^3}{27}}}{2} = \frac{c}{2} \pm \sqrt{\left(\frac{c}{2}\right)^2 + \left(\frac{b}{3}\right)^3}$$

(8)  From the symmetry of the equation, we can assume that:

$$u^3 = \frac{c}{2} + \sqrt{\left(\frac{c}{2}\right)^2 + \left(\frac{b}{3}\right)^3}$$

$$v^3 = c - u^3 = c - \frac{c}{2} - \sqrt{\left(\frac{c}{2}\right)^2 + \left(\frac{b}{3}\right)^3} = \frac{c}{2} - \sqrt{\left(\frac{c}{2}\right)^2 + \left(\frac{b}{3}\right)^3}$$

(9)  $x = u + v = \sqrt[3]{\frac{c}{2} + \sqrt{\left(\frac{c}{2}\right)^2 + \left(\frac{b}{3}\right)^3}} + \sqrt[3]{\frac{c}{2} - \sqrt{\left(\frac{c}{2}\right)^2 + \left(\frac{b}{3}\right)^3}}$