Bézout method

The Bézout Method is a general method of solving algebraic equations. It was initiated and developed by Étienne Bézout in 1762.

This method tries to reduce the equation that we want to solve to other equations of lower degree. This tedious method certainly fails for equations of degree greater than or equal to five that have an unresolved Galois group. It has a concrete interest only for equations of degree 3.

Principle of the method

Consider an equation of degree n :

$\qquad a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}=0$

Let r be an n -th primitive root of unity.

We know that the n roots n -ths of unity 1, r , r ² , ..., r ^{n − 1} verify the relation:

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

Bézout's method is looking for the roots of the studied equation in the form of linear combinations of the roots n-ths of the unit.

$x=b_{0}+b_{1}r+b_{2}r^{2}+\cdots +b_{n-1}r^{n-1}$

For this, we start by eliminating r between the two relations:

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

$x=b_{0}+b_{1}r+b_{2}r^{2}+\cdots +b_{n-1}r^{n-1}$

This gives us an equation of degree n in x whose coefficients are expressions depending on b₀, b₁, b₂,...,b_n−1. By identifying the coefficients of this equation with the corresponding coefficients of the equation to be solved, we obtain a system of equations of unknowns b₀, b₁, b₂,...,b_n−1 which after solving and reporting the different solutions in:

$x=b_{0}+b_{1}r+b_{2}r^{2}+\cdots +b_{n-1}r^{n-1}$

will give us the solutions of the equation that we had set ourselves to solve.

Application to the resolution of cubic equations

We will expose the method on the following example:

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

Let

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

j is one of the cube roots of the unit and therefore satisfies:

$\mathrm {j} ^{3}=1~$

We will search for the roots in the form:

$x=a+b\mathrm {j} +c\mathrm {j} ^{2}$

We will eliminate j between the last two equations.

The last two equations are in the form:

$\left\{{\begin{matrix}\mathrm {j} ^{3}=1\\x-a-b\mathrm {j} =c\mathrm {j} ^{2}\end{matrix}}\right.$

By making successive member-to-member products and each time replacing those of the two equations whose degree with respect to j is the highest by the result, we will gradually lower the degree of the equations with respect to j until j disappears from one of the equations.

A first member-to-member product gives us:

$\left\{{\begin{matrix}b\mathrm {j} ^{2}=\mathrm {j} x-a\mathrm {j} -c\\x-a-b\mathrm {j} =c\mathrm {j} ^{2}\end{matrix}}\right.$

A second member-to-member product gives us:

$\left\{{\begin{matrix}c\mathrm {j} x-ac\mathrm {j} +b^{2}\mathrm {j} =bx+c^{2}-ab\\x-a-b\mathrm {j} =c\mathrm {j} ^{2}\end{matrix}}\right.$

A third member-to-member product gives us:

$\left\{{\begin{matrix}c\mathrm {j} x-ac\mathrm {j} +b^{2}\mathrm {j} =bx+c^{2}-ab\\ab^{2}-a^{2}c-b^{2}x+2acx-cx^{2}=2abc\mathrm {j} -b^{3}\mathrm {j} -c^{3}\mathrm {j} -2bc\mathrm {j} x\end{matrix}}\right.$

A last member-to-member product eliminates j and gives 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 we need to solve, we obtain:

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

From the first equation we deduce from this the value of a that we report in the other equations, we obtain:

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

Let's memorize the value of a and bring the product bc in the third equation, we get:

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

By cubing the two members of the first equation, we obtain:

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

b ³ and c ³ are therefore 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 pairs ( b , c ) checking:

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

thereby are :

$b_{1}={\frac {1}{3}}{\sqrt[{3}]{\frac {5}{2}}}\quad {\text{و}}\quad c_{1}=-{\frac {1}{3}}{\sqrt[{3}]{50}};$

$b_{2}={\frac {\mathrm {j} }{3}}{\sqrt[{3}]{\frac {5}{2}}}\quad {\text{و}}\quad c_{2}=-{\frac {\mathrm {j} ^{2}}{3}}{\sqrt[{3}]{50}};$

$b_{3}={\frac {\mathrm {j} ^{2}}{3}}{\sqrt[{3}]{\frac {5}{2}}}\quad {\text{و}}\quad c_{3}=-{\frac {\mathrm {j} }{3}}{\sqrt[{3}]{50}}.$

By reporting in $x=a+b\mathrm {j} +c\mathrm {j} ^{2}$ the values ​​of $a,b,c$ found, we get

$x_{1}={\frac {1}{3}}+{\frac {1}{3}}{\sqrt[{3}]{\frac {5}{2}}}\mathrm {j} -{\frac {1}{3}}{\sqrt[{3}]{50}}\mathrm {j} ^{2},$

$x_{2}={\frac {1}{3}}+{\frac {\mathrm {j} }{3}}{\sqrt[{3}]{\frac {5}{2}}}\mathrm {j} -{\frac {\mathrm {j} ^{2}}{3}}{\sqrt[{3}]{50}}\mathrm {j} ^{2}$

$x_{3}={\frac {1}{3}}+{\frac {\mathrm {j} ^{2}}{3}}{\sqrt[{3}]{\frac {5}{2}}}j-{\frac {j}{3}}{\sqrt[{3}]{50}}\mathrm {j} ^{2},$

which, after simplification, gives

$x_{1}={\frac {1}{3}}\left(1+\mathrm {j} {\sqrt[{3}]{\frac {5}{2}}}-\mathrm {j} ^{2}{\sqrt[{3}]{50}}\right),$

$x_{2}={\frac {1}{3}}\left(1+\mathrm {j} ^{2}{\sqrt[{3}]{\frac {5}{2}}}-\mathrm {j} {\sqrt[{3}]{50}}\right)$

$x_{3}={\frac {1}{3}}\left(1+{\sqrt[{3}]{\frac {5}{2}}}-{\sqrt[{3}]{50}}\right)$,

which are the three roots of the equation that we had to solve.

External link

• Text Bézout (1764) on the resolution of algebraic equations, online and commented on BibNum

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