The Cubic Formula

Let $b,c,d$ be constants with $a \neq 0$ .

In this post, we develop a formula to solve the cubic equation

$x^3 + bx^2 + cx + d = 0.$

Problem 1. Using the substitution $x = y + h$ , determine the constants $B,C,D$ in terms of $b,c,d,h$ such that

$x^3 + bx^2 + cx + d = y^3 + By^2 + Cy + D.$

Deduce the value of $h$ such that the right-hand side has no $y^2$ term.

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Solution. Making the substitution on the left-hand side,

$\begin{aligned} x^3 + bx^2 + cx + d &= (y+h)^3 + b(y+h)^2 \\ &\phantom{=.} + c(y+h) + d \\ &= (y^3 + 3y^2h + 3yh^2 + h^3) \\ &\phantom{=.} + b(y^2 + 2yh + h^2) \\ &\phantom{=.} + c(y+h) + d \\ &= y^3 + (3h + b)y^2 \\ &\phantom{=.} + (3h^2 + 2bh + c)y \\ &\phantom{=.} + (h^3 + bh^2 + ch + d). \end{aligned}$

Therefore,

$\begin{aligned} B &= 3h + b, \\ C &= 3h^2 + 2bh + c, \\ D &= h^3 + bh^2 + ch + d. \end{aligned}$

Hence, we set $B = 0 \iff h = -b/3$ .

Remark 1. Substituting $x = y - b/3$ ,

$x^3 + bx^2 + cx + d = 0 \quad \iff \quad y^3 + Cy + D = 0.$

Therefore, solving the right-hand side, called the depressed cubic, allows us to solve the left-hand side.

Problem 2 (Viète’s Substitution). Making the substitution

$\displaystyle y = z - \frac C{3z}$

into the right-hand side, obtain an equation that is quadratic in $z^3$ .

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Solution. By performing painful algebra,

$\begin{aligned} y^3 + Cy &= y(y^2 + C) \\ &= \left( z - \frac C{3z} \right) \left( \left( z - \frac C{3z} \right)^2 + C \right) \\ &= \left( z - \frac C{3z} \right) \left( z^2 - 2 \cdot z \cdot \frac{C}{3z} + \frac{C^2}{(3z)^2} + C \right) \\ &= \left( z - \frac C{3z} \right) \left( z^2 + \frac{C^2}{9z^2} + \frac C3 \right) \\ &= z^3 -\frac{C^3}{27z^3}. \end{aligned}$

Therefore,

$\begin{aligned} y^3 + Cy + D &= 0\\ z^3 -\frac{C^3}{27z^3} + D &= 0 \\ (z^3)^2 + Dz^3 - \frac{C^3}{27} &= 0. \end{aligned}$

Assume $D^2/4 + C^3/27 \geq 0$ .

Problem 3. Using Problems 1 and 2, solve the equation

$x^3 + bx^2 + cx + d = 0.$

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Solution. Using Problem 2, we solve for $z^3$ using the quadratic formula: $m := -D/2$ and $p := -C^2/27$ , so that

$\begin{aligned} m^2 - p &= \left( -\frac D2 \right)^2 - \left( - \frac{C^3}{27} \right) \\ &= \frac{D^2}{4} + \frac{C^2}{27}. \end{aligned}$

yields

$\begin{aligned} z^3 &= m \pm \sqrt{m^2 - p} \\ z^3 &= -\frac D2 \pm \sqrt{\frac{D^2}{4} + \frac{C^3}{27}} \\ z_{\pm} &:= \sqrt[3]{ -\frac D2 \pm \sqrt{\frac{D^2}{4} + \frac{C^3}{27}} }. \end{aligned}$