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

##### 2 Answers

#### Answer:

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

**Discriminant**

The discriminant

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

In our example,

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

Since

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

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

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

Multiply through by

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

Note that this is of the form

#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_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:

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

#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

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#