Question

How do you solve `x^3-3x+1=0` using Cadano's method ?

Answer 1

See explanation...

Explanation:

Assuming you mean Cardano's method, I would remark that it is not a good choice for this particular cubic. This cubic has `3` real zeros, as we can find by examining its discriminant:

`color(white)()`
Discriminant

The discriminant `Delta` of a cubic polynomial in the form `ax^3+bx^2+cx+d` is given by the formula:

`Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd`

In our example, `a=1`, `b=0`, `c=-3` and `d=1`, so we find:

`Delta = 0+108+0-27+0 = 81`

Since `Delta > 0` this cubic has `3` Real zeros.

As a result, if we attempt to solve using Cardano's method then the solution will be expressed in terms of irreducible cube roots of complex numbers - Cardano's "casus irreducibilis".

Let's go ahead anyway to see it happen...

Given:

`x^3-3x+1 = 0`

Let `x = u+v`.

Then:

`(u+v)^3-3(u+v)+1 = 0`

which expands to:

`u^3+v^3+3(uv-1)(u+v)+1 = 0`

Add the constraint `v = 1/u` to eliminate the `(u+v)` term and get:

`u^3+1/u^3+1 = 0`

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

`(u^3)^2+(u^3)+1 = 0`

Note that this is of the form `t^2+t+1 = 0`, which if multiplied by `(t-1)` gives `t^3-1 = 0`. In other words, its solutions are the non-real complex cube roots of `1`:

`omega = -1/2+sqrt(3)/2i = cos((2pi)/3) + i sin((2pi)/3)`

`omega^2 = -1/2-sqrt(3)/2i = cos(-(2pi)/3) + i sin(-(2pi)/3)`

Then using `x=u+v` and `v=1/u`, we find solutions of our cubic:

`x_1 = root(3)(omega)+1/root(3)(omega) = 2cos((2pi)/9) ~~ 1.5321`

`x_2 = omega root(3)(omega)+ 1/(omega root(3)omega) = 2cos((8pi)/9) ~~ -1.8794`

`x_3 = omega^2 root(3)(omega) + 1/(omega^2 root(3)(omega)) = 2cos((14pi)/9) ~~ 0.34730`

Answer 2

Here's an alternative method...

Explanation:

Given:

`x^3-3x+1 = 0`

Here's an alternative trigonometric sustitution method of solution, suitable for cubics such as this one, with three real zeros (Cardano's "casus irreducibilis"):

Consider the substitution:

`x = k cos theta`

Then our cubic equation becomes:

`0 = (k cos theta)^3-3(k cos theta)+1`

`color(white)(0) = k(k^2 cos^3 theta - 3 cos theta) + 1`

Putting `k=2` this becomes:

`0 = 2(4 cos^3 theta - 3 cos theta) + 1`

`color(white)(0) = 2cos 3 theta + 1`

Hence:

`cos 3 theta = -1/2`

Hence:

`3 theta = +- cos^(-1)(-1/2) + 2kpi = +-(2pi)/3+2kpi`

for any integer `k`.

So:

`theta = +-((2+6k)pi)/9`

So:

`x = 2 cos (+-((2+6k)pi)/9)`

which results in distinct values:

`x_1 = 2 cos ((2pi)/9) ~~ 1.53209`

`x_2 = 2 cos ((8pi)/9) ~~ -1.87939`

`x_3 = 2 cos ((14pi)/9) ~~ 0.34730`