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

##### 1 Answer

#### Answer:

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

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

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

Substitute

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

Let

#t^3+pt+q = 0#

Next substitute

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

Then add the constraint:

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

Multiply through by

#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

If

#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

If on the other hand

In this case to match the appropriate cube root

#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))#

From these vales of