Question

How do you solve standard cubic equations, and is there a general formula like there is for quadratics?

Answer 1

It's a little complicated...

Explanation:

There are standard formulas, but they are somewhat messy and I would not recommend learning them.

Here's a sequence of steps for solving a general cubic equation:

Given:

`ax^3+bx^2+cx+d = 0" "` with `a != 0`

Divide the whole equation by `a` to get:

`x^3+b/ax^2+c/ax+d/a = 0`

Note that:

`(x+b/(3a))^3 = x^3+b/ax^2+b^2/(3a)x+b^3/(27a^3)`

`(c/a-b^2/(3a))(x+b/(3a)) = (c/a-b^2/(3a))x+((bc)/(3a^2)-b^3/(9a^2))`

So substitute `t = x+b/(3a)` to get:

`t^3+(c/a-b^2/(3a))t+(d/a+b^3/(9a^2)-(bc)/(3a^2)-b^3/(27a^3)) = 0`

Let:

`p = (c/a-b^2/(3a))`

`q = (d/a+b^3/(9a^2)-(bc)/(3a^2)-b^3/(27a^3))`

So:

`t^3+pt+q = 0`

Next let:

`t = u+v`

So our equation becomes:

`u^3+v^3+(3uv+p)(u+v) + q = 0`

To eliminate the term in `(u+v)` add the constraint:

`3uv+p = 0`

Hence:

`u^3-p^3/(27u^3)+q = 0`

Multiplying through by `9u^3` this becomes:

`9(u^3)^2+9q(u^3)-p^3/3 = 0`

Then by the quadratic formula we find:

`u^3 = (-9q+-sqrt(81q^2+12p^3))/18`

This would give `6` possible values for `u`, but they are also values for `v` and there are three distinct resulting sums. So we can take the positive sign of the `+-` here and find...

Hence three roots:

`u_k = omega^k root(3)((-9q+sqrt(81q^2+12p^3))/18)" "k = 0,1,2`

where `omega = -1/2+sqrt(3)/2i` is the primitive complex cube root of `1`

Then `v_k = -p/(3u_k)`

So three roots to our reduced cubic:

`t_k = omega^k root(3)((-9q+sqrt(81q^2+12p^3))/18) - (omega^(-k) p)/(3root(3)((-9q+sqrt(81q^2+12p^3))/18))`

and hence three roots to our original cubic:

`x_k = -b/(3a)+omega^k root(3)((-9q+sqrt(81q^2+12p^3))/18) - (omega^(-k) p)/(3root(3)((-9q+sqrt(81q^2+12p^3))/18))" "k = 0,1,2`

Actually, if exactly one of the roots of the cubic is real, then we can simplify this to:

`x_k = -b/(3a)+omega^k root(3)((-9q+sqrt(81q^2+12p^3))/18)+omega^(-k)root(3)((-9q-sqrt(81q^2+12p^3))/18) " "k = 0,1,2`

This derivation works ok when `81q^2+12p^3 >= 0`, but gives a nasty formulation when `81q^2+12p^3 < 0`. In particular the real root is described in terms of irreducible cube roots of complex numbers.

As an alternative, if the cubic has `3` real roots we can use a trigonometric substitution `t = k cos theta` to get a result expressed in terms of `cos` and `cos^(-1)`.

Answer 2

Here's the trigonometric method for the case where the cubic equation has `3` real roots...

Explanation:

Let us deal with the reduced cubic:

`t^3+pt+q = 0`

(see other solution for how to get here if your cubic is more general)

Consider the substitution:

`t = k cos theta`

Then our equation becomes:

`k^3 cos^3 theta + kp cos theta + q = 0`

We want to use the trigonometric identity:

`cos 3 theta = 4 cos^3 theta - 3 cos theta`

So dividing our equation by `k^3/4` we find:

`4 cos^3 theta + (4p)/k^2 cos theta + (4q)/k^3 = 0`

Then we want:

`(4p)/k^2 = -3`

So:

`k^2 = -(4p)/3`

and we can choose:

`k = sqrt(-(4p)/3) = 2/3sqrt(-3p)`

Note also:

`k^4 = (-(4p)/3)^2 = (16p)/9`

Then:

`(4q)/k^3 = (4qk)/k^4 = (4qk)/((16p)/9) = (9qk)/4 = 3/2qsqrt(-3p)`

So our equation becomes:

`4 cos^3 theta-3 cos theta + 3/2qsqrt(-3p) = 0`

That is:

`cos 3 theta = -3/2qsqrt(-3p)`

So:

`3 theta = +-cos^(-1)(-3/2qsqrt(-3p)) + 2npi" "n in ZZ`

Hence distinct values:

`cos theta = cos(1/3 cos^(-1)(-3/2qsqrt(-3p))+(2npi)/3)" "n = 0, 1, 2`

Then `t = k cos theta`, so the distinct roots of our original cubic are:

`t_n = 2/3sqrt(-3p) cos(1/3 cos^(-1)(-3/2qsqrt(-3p))+(2npi)/3)" "n = 0, 1, 2`

Footnote

If not all of the roots are real, then `abs(-3/2qsqrt(-3p)) > 1`. The formula we derived is still true, but we have to deal with complex values of `cos^(-1)`.

Alternatively, note that `cosh theta` also satisfies:

`cosh 3 theta = 4 cosh^3 theta - 3 cosh theta`

So we can follow a similar derivation to find:

`t_n = 2/3sqrt(-3p) cosh(1/3 cosh^(-1)(-3/2qsqrt(-3p))+(2npi)/3i)" "n = 0, 1, 2`

I have not really used this hyperbolic method, since in the cases where it makes most sense we can use Cardano's method anyway.