Method of Bézout

The method of Bezout , imagined and developed by Etienne Bézout in 1762, is a general method of resolution of the algebraic equations.

This method tries to bring back the equation which one wants to solve with other equations of less low degree. This method fails in an unquestionable way for the equations of degrees equal to or higher than five which have a Groupe of nonresolvable Welshman.

This method, tiresome for the equation S of degree equal to or higher than 4, has a concrete interest only for the equations of degree 3.

Principle of the method

Let us consider an equation of degree N:

$\ qquad a_n x^n + a_ {N - 1} x^ {N - 1} + \ cdots + a_1 X + a_0 = 0$

That is to say R n-ème a primary root of the unit.

We know that N n-ème roots of the 1, r unit, r²,…, r^n-1 check the relation:

$\ qquad 1 + R + r^2 + \ cdots + r^ {n-1} = 0$

The method of Bezout consists in seeking the roots of the equation studied in the form of linear combinations of the n-ème roots of the unit.

$\ qquad X = b_0 + b_ 1r + b_ 2r ^2 + \ cdots + b_ {n-1} r^ {n-1}$

For that, one starts by eliminating R between the two relations:

$\ qquad 1 + R + r^2 + \ cdots + r^ {n-1} = 0$

$\ qquad X = b_0 + b_ 1r + b_ 2r ^2 + \ cdots + b_ {n-1} r^ {n-1}$

What gives us an equation of degree N in X whose coefficients are expressions depending on b₀, b₁, b₂,…, b_n. By identifying the coefficients of this equation with the coefficients corresponding of the equation to solve, one obtains a system of equations of unknown factors b₀, b₁, b₂,…, b_n which after resolution and carryforward of the various solutions in:

$\ qquad X = b_0 + b_ 1r + b_ 2r ^2 + \ cdots + b_ {n-1} r^ {n-1}$

we will give the solutions of the equation which one had been given to solve.

Application to the resolution of the cubic equations

We will expose the method on the following example:

$6x^3 - 6x^2 + 12x + 7=0 ~$

Let us pose:

$J = e^ {\ frac {2i \ pi} {3}} ~$

J is one of the cubic roots of the unit and thus checks:

$j^3 = 1 ~$

Let us seek the roots in the form:

$X = a+bj+cj^2 \ qquad (*) ~$

We will eliminate J between the two last equations.

The two last equations are put in the form:

$\ left \ {\ begin {matrix} j^3=1 \ \ x-a-bj=cj^2 \ end {matrix} \ right.$

While making products successive member with member and while replacing each time that of the two equations of which the degrees compared to J is raised the most by the result, we will gradually lower the degree of the equations compared to J until J disparraisse one of the equations.

A first product member with member gives us:

$\ left \ {\ begin {matrix} bj^2=jx-aj-c \ \ x-a-bj=cj^2 \ end {matrix} \ right.$

A second product member with member gives us:

$\ left \ {\ begin {matrix} cjx-acj+b^2j=bx+c^2-ab \ \ x-a-bj=cj^2 \ end {matrix} \ right.$

A third product member with member gives us:

$\ left \ {\ begin {matrix} cjx-acj+b^2j=bx+c^2-ab \ \ ab^2-a^2c-b^2x+2acx-cx^2=2abcj-b^3j-c^3j-2bcjx \ end {matrix} \ right.$

A last product member with member makes it possible to eliminate J and provides us the equation:

$x^3 - 3ax^2 + (3a^2 - 3bc) X + 3abc - a^3 - b^3 - c^3 = 0 ~$

By identifying the coefficients of this equation with the coefficients of the equation which we must solve, we obtain:

$\ left \ {\ begin {matrix} -3a=-1 \ \ 3a^2 - 3bc=2 \ \ 3abc - a^3 - b^3 - c^3= \ frac {7} {6} \ end {matrix} \ right.$

First equation we let us deduce the value from it from has that one defers in the other equations, one obtains:

$\ left \ {\ begin {matrix} a= \ frac {1} {3} \ \ bc=- \ frac {5} {9} \ \ bc - b^3 - c^3= \ frac {65} {54} \ end {matrix} \ right.$

The value of has and carry the product bc in the third equation, we memorize obtain:

$\ left \ {\ begin {matrix} bc=- \ frac {5} {9} \ \ b^3 + c^3 = - \ frac {95} {54} \ end {matrix} \ right.$

By raising with the cube the two members of the first equation, one obtains:

$\ left \ {\ begin {matrix} b^3c^3 = - \ frac {125} {729} \ \ b^3+c^3 = - \ frac {95} {54} \ end {matrix} \ right. ~$

b³ and c³ are thus the roots of the equation:

$X^2 + \ frac {95} {54} X - \ frac {125} {729} = 0 ~$

The two roots of this equation are:

$b^3 = \ frac {5} {54} ~$

$c^3 = - \ frac {50} {27} ~$

The three couples (B, c) checking:

$bc = - \ frac {5} {9} ~$

are thus:

$b_1 = \ frac {1} {3} \ sqrt {\ frac {5} {2}}$ and $\ qquad c_1 = - \ frac {1} {3} \ sqrt {50}$

$b_2 = \ frac {J} {3} \ sqrt {\ frac {5} {2}}$ and $\ qquad c_2 = - \ frac {j^2} {3} \ sqrt {50}$

$\ qquad b_3 = \ frac {j^2} {3} \ sqrt {\ frac {5} {2}}$ and $\ qquad c_3 = - \ frac {J} {3} \ sqrt {50}$

While deferring in (*) the values of has, B, C found, one obtains:

$x_1 = \ frac {1} {3} + \ frac {1} {3} \ sqrt {\ frac {5} {2}} J - \ frac {1} {3} \ sqrt {50} j^2 ~$

$x_2 = \ frac {1} {3} + \ frac {J} {3} \ sqrt {\ frac {5} {2}} J - \ frac {j^2} {3} \ sqrt {50} j^2 ~$

$x_3 = \ frac {1} {3} - \ frac {j^2} {3} \ sqrt {50} j+ \ frac {J} {3} \ sqrt {\ frac {5} {2}} j^2 ~$

Who, after simplification gives:

$x_1 = \ frac {1} {3} (1 + J \ sqrt {\ frac {5} {2}} - j^2 \ sqrt {50}) ~$

$x_2 = \ frac {1} {3} (1 + j^2 \ sqrt {\ frac {5} {2}} - J \ sqrt {50}) ~$

$x_3 = \ frac {1} {3} (1 - \ sqrt {50} + \ sqrt {\ frac {5} {2}}) ~$

Who are the three roots of the equation which one was to solve.

Method of Bézout

Principle of the method

Application to the resolution of the cubic equations

Other methods of solution of equations