Question

What is the solution for a cubic equation? Thanks.....

Answer 1

See explanation for a demonstration of Cardano's method...

Explanation:

There is a general formula and there are several methods.

My personal favourite method, which works well when a cubic equation has one Real root and two non-Real roots is Cardano's method.

Given a cubic equation:

`ax^3+bx^2+cx+d = 0`

If the coefficients are integers, then it may be useful to first multiply through by `3^3 a^2` to reduce the number of fractions we need to deal with.

`27a^3x^3 + 27a^2bx^2 + 27a^2cx+27a^2d=0`

`(3ax)^3+3b(3ax)^2+9ac(3ax)+27a^2d = 0`

Substitute `t = 3ax+b` to get:

`t^3+(9ac-3b^2)t+(27a^2d+2b^3-9abc) = 0`

Let `p = 9ac-3b^2` and `q = 27a^2d+2b^3-9abc`, so we have:

`t^3+pt+q = 0`

Next substitute `t = u + v` to get:

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

Then add the constraint: `v = -p/(3u)` to make `(3uv+p) = 0`

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

Multiply through by `27u^3` and rearrange to get:

`27(u^3)^2+27q(u^3)-p^3 = 0`

Use the quadratic formula to find:

`u^3 = (-27q+-sqrt(729q^2+108p^3))/54`

`=(-27q+-3sqrt(81q^2+12p^3))/54`

The derivation was symmetric in `u` and `v`, so one of these roots can be used for `u^3` and the other for `v^3`.

If `81q^2+12p^3 >= 0` then the square root is Real and the Real root of the cubic `t^3+pt+q = 0` is:

`t_1 = root(3)((-27q+3sqrt(81q^2+12p^3))/54)+root(3)((-27q-3sqrt(81q^2+12p^3))/54)`

In this case, the Complex roots are given by:

`t_2 = omega root(3)((-27q+3sqrt(81q^2+12p^3))/54)+omega^2 root(3)((-27q-3sqrt(81q^2+12p^3))/54)`

`t_3 = omega^2 root(3)((-27q+3sqrt(81q^2+12p^3))/54)+omega root(3)((-27q-3sqrt(81q^2+12p^3))/54)`

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

If on the other hand `81q^2+12p^3 < 0` then the square roots are pure imaginary, the cube roots are of Complex numbers and all three roots are Real.

In this case to match the appropriate cube root `v` to `u`, we have to use the relation `v = -p/(3u)` to find:

`t_(k+1) = omega^k root(3)((-27q+3sqrt(81q^2+12p^3))/54) - p/(3 omega^k root(3)((-27q+3sqrt(81q^2+12p^3))/54))`

`k = 0, 1, 2`

From these vales of `t` we can find `x` by `x = (t-b)/(3a)`