First the cubic is depressed, by using the Tschirnhaus Transformation: :$x \to x + \dfrac b {3 a}$: {{begin-eqn}} {{eqn | l = a x^3 + b x^2 + c x + d       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x + \frac b {3 a} }^3 - 3 \frac b {3 a} x^2 - 3 \frac {b^2} {9 a^2} x - \frac {b^3} {27 a^3} + \frac b a x^2 + \frac c a x + \frac d a       | r = 0       | c = completing the cube: $\paren {a + b}^3 = a^3 + 3 a^2 b + 3 a b^2 + b^3$ }} {{eqn | ll= \leadsto       | l = \paren {x + \frac b {3 a} }^3 + \paren {\frac {3 a c - b^2} {3 a^2} } x + \paren {\frac {27 a^2 d - b^3} {27 a^3} }       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x + \frac b {3 a} }^3 + \paren {\frac {3 a c - b^2} {3 a^2} } \paren {x + \frac b {3 a} } - \frac b {3 a} \paren {\frac {3 a c - b^2} {3 a^2} } + \paren {\frac {27 a^2 d - b^3} {27 a^3} }       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x + \frac b {3 a} }^3 + \paren {\frac {3 a c - b^2} {3 a^2} } \paren {x + \frac b {3 a} } + \paren {\frac {27 a^2 d + 2 b^3 - 9 a b c} {27 a^3} }       | r = 0       | c =  }} {{end-eqn}} Now let: :$y = x + \dfrac b {3 a}, Q = \dfrac {3 a c - b^2} {9 a^2}, R = \dfrac {9 a b c - 27 a^2 d - 2 b^3} {54 a^3}$ Thus we have obtained the depressed cubic $y^3 + 3 Q y - 2 R = 0$. Now let $y = u + v$ where $u v = -Q$. Then: {{begin-eqn}} {{eqn | l = \paren {u + v}^3 + 3 Q \paren {u + v} - 2 R       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = u^3 + 3 u^2 v + 3 u v^2 + v^3 + 3 Q \paren {u + v} - 2 R       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = u^3 - 3 Q \paren {u + v} + 3 Q \paren {u + v} + v^3 - 2 R       | r = 0       | c = as $u v = -Q$ }} {{eqn | ll= \leadsto       | l = u^3 + v^3 - 2 R       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = u^3 - \frac {Q^3} {u^3} - 2 R       | r = 0       | c = substituting for $T$ from $u v = -Q$ }} {{eqn | ll= \leadsto       | l = u^6 - 2 R u^3 - Q^3       | r = 0       | c =  }} {{end-eqn}} Thus the resolvent equation is obtained. This resolvent is seen to be a quadratic in $u^3$. From Solution to Quadratic Equation: :$u^3 = \dfrac {2 R \pm \sqrt {4 Q^3 + 4 R^2}} 2 = R \pm \sqrt {Q^3 + R^2}$ We have from above $u v = -Q$ and hence $v^3 = -\dfrac {Q^3} {u^3}$. Let us try taking the positive root: $u^3 = R + \sqrt {Q^3 + R^2}$. Then: {{begin-eqn}} {{eqn | l = v^3       | r = \frac {-Q^3} {R + \sqrt {Q^3 + R^2} }       | c =  }} {{eqn | r = \frac {-Q^3 \paren {R - \sqrt {Q^3 + R^2} } } {R^2 - \paren {Q^3 + R^2} }       | c =  }} {{eqn | r = R - \sqrt {Q^3 + R^2}       | c =  }} {{end-eqn}} The same sort of thing happens if you start with $u^3 = R - \sqrt {Q^3 + R^2}$: we get $v^3 = R + \sqrt {Q^3 + R^2}$. Thus we see that taking either square root we arrive at the same solution. {{WLOG}}: :$u^3 = R + \sqrt {Q^3 + R^2}$ :$v^3 = R - \sqrt {Q^3 + R^2}$ Let: :$S = \sqrt [3] {R + \sqrt {Q^3 + R^2} }$ :$T = \sqrt [3] {R - \sqrt {Q^3 + R^2} }$ From Roots of Complex Number, we have the three cube roots of $u^3$ and $v^3$: :$u = \begin{cases} & S \\ \paren {-\dfrac 1 2 + \dfrac {i \sqrt 3} 2} & S \\  \paren {-\dfrac 1 2 - \dfrac {i \sqrt 3} 2} & S \\  \end{cases}$ :$v = \begin{cases} & T \\ \paren {-\dfrac 1 2 + \dfrac {i \sqrt 3} 2} & T \\  \paren {-\dfrac 1 2 - \dfrac {i \sqrt 3} 2} & T \\  \end{cases}$ Because of our constraint $u v = -Q$, there are only three combinations of these which are possible such that $y = u + v$: :$ y = \begin{cases} & S + T  \\ \paren {-\dfrac 1 2 + \dfrac {i \sqrt 3} 2} S + \paren {-\dfrac 1 2 - \dfrac {i \sqrt 3} 2} T = & -\dfrac {S + T} 2 + \dfrac {i \sqrt 3} 2 \paren {S - T} \\  \paren {-\dfrac 1 2 - \dfrac {i \sqrt 3} 2} S + \paren {-\dfrac 1 2 + \dfrac {i \sqrt 3} 2} T = & -\dfrac {S + T} 2 - \dfrac {i \sqrt 3} 2 \paren {S - T} \\  \end{cases}$ As $y = x + \dfrac b {3a}$, it follows that the three roots are therefore: :$(1): \quad x_1 = S + T - \dfrac b {3 a}$ :$(2): \quad x_2 = - \dfrac {S + T} 2 - \dfrac b {3 a} + \dfrac {i \sqrt 3} 2 \paren {S - T}$ :$(3): \quad x_3 = - \dfrac {S + T} 2 - \dfrac b {3 a} - \dfrac {i \sqrt 3} 2 \paren {S - T}$ {{qed}}