Question Number 54775 by MJS last updated on 10/Feb/19
![found something interesting, it was published by Tschirnhaus in 1683 we can reduce x^3 +ax^2 +bx+c=0 (1) to y^3 +py+q=0 (2) and further to z^3 =t (1) is the well known linear substitution y=x+(a/3) → x=y−(a/3) ⇒ y^3 −((a^2 −3b)/3)y+((2a^3 −9ab+27c)/(27))=0 p=−((a^2 −3b)/3) and q=((2a^3 −9ab+27c)/(27)) ⇒ y^3 +py+q=0 (2) quadratic substitution z=y^2 +αy+β → y^2 +αy+(β−z)=0 we could solve this for y and then plug in above... (Tschirnhaus did) but there′s an easier way: we calculate the determinant of the Sylvester Matrix we have (a) 1y^3 +0y^2 +py+q=0 (b) 0y^3 +y^2 +αy+(β−z)=0 the matrix is [(1,0,p,q,0),(0,1,0,p,q),(0,1,α,(β−z),0),(0,0,1,α,(β−z)),(1,α,(β−z),0,0) ] the determinant is −z^3 +(3β−2p)z^2 −(pα^2 +3qα+3β^2 −4pβ+p^2 )z −(qα^3 −pα^2 β−3qαβ+pqα−β^3 +2pβ^2 −p^2 β−q^2 )p we want the square and the linear terms to disappear so we set their constants zero to get α and β ⇒ β=((2p)/3); α=−((3q)/(2p))±((√(12p^3 +81q^2 ))/(6p)) this leads to z^3 =((8p^3 )/(27))+((27q^4 )/(2p^3 ))+4q^2 ±((q(√(3(4p^3 +27q^2 )^3 )))/(18p^3 )) Tschirnhaus thought he could solve polynomes of any degree with this method but it′s getting harder to solve because you need a cubic substitution to eliminate 3 constants and so on...](https://www.tinkutara.com/question/Q54775.png)
$$\mathrm{found}\:\mathrm{something}\:\mathrm{interesting},\:\mathrm{it}\:\mathrm{was}\:\mathrm{published} \\ $$$$\mathrm{by}\:\mathrm{Tschirnhaus}\:\mathrm{in}\:\mathrm{1683} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{reduce} \\ $$$${x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{to} \\ $$$${y}^{\mathrm{3}} +{py}+{q}=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{and}\:\mathrm{further}\:\mathrm{to} \\ $$$${z}^{\mathrm{3}} ={t} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{well}\:\mathrm{known}\:\mathrm{linear}\:\mathrm{substitution} \\ $$$${y}={x}+\frac{{a}}{\mathrm{3}}\:\rightarrow\:{x}={y}−\frac{{a}}{\mathrm{3}} \\ $$$$\Rightarrow\:{y}^{\mathrm{3}} −\frac{{a}^{\mathrm{2}} −\mathrm{3}{b}}{\mathrm{3}}{y}+\frac{\mathrm{2}{a}^{\mathrm{3}} −\mathrm{9}{ab}+\mathrm{27}{c}}{\mathrm{27}}=\mathrm{0} \\ $$$${p}=−\frac{{a}^{\mathrm{2}} −\mathrm{3}{b}}{\mathrm{3}}\:\mathrm{and}\:{q}=\frac{\mathrm{2}{a}^{\mathrm{3}} −\mathrm{9}{ab}+\mathrm{27}{c}}{\mathrm{27}} \\ $$$$\Rightarrow\:{y}^{\mathrm{3}} +{py}+{q}=\mathrm{0} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:\mathrm{quadratic}\:\mathrm{substitution} \\ $$$${z}={y}^{\mathrm{2}} +\alpha{y}+\beta\:\rightarrow\:{y}^{\mathrm{2}} +\alpha{y}+\left(\beta−{z}\right)=\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{could}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{for}\:{y}\:\mathrm{and}\:\mathrm{then}\:\mathrm{plug}\:\mathrm{in} \\ $$$$\mathrm{above}…\:\left(\mathrm{Tschirnhaus}\:\mathrm{did}\right)\:\mathrm{but}\:\mathrm{there}'\mathrm{s}\:\mathrm{an} \\ $$$$\mathrm{easier}\:\mathrm{way}:\:\mathrm{we}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{determinant}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{Sylvester}\:\mathrm{Matrix} \\ $$$$\mathrm{we}\:\mathrm{have} \\ $$$$\left({a}\right)\:\mathrm{1}{y}^{\mathrm{3}} +\mathrm{0}{y}^{\mathrm{2}} +{py}+{q}=\mathrm{0} \\ $$$$\left({b}\right)\:\mathrm{0}{y}^{\mathrm{3}} +{y}^{\mathrm{2}} +\alpha{y}+\left(\beta−{z}\right)=\mathrm{0} \\ $$$$\mathrm{the}\:\mathrm{matrix}\:\mathrm{is} \\ $$$$\begin{bmatrix}{\mathrm{1}}&{\mathrm{0}}&{{p}}&{{q}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}&{{p}}&{{q}}\\{\mathrm{0}}&{\mathrm{1}}&{\alpha}&{\beta−{z}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}&{\alpha}&{\beta−{z}}\\{\mathrm{1}}&{\alpha}&{\beta−{z}}&{\mathrm{0}}&{\mathrm{0}}\end{bmatrix} \\ $$$$\mathrm{the}\:\mathrm{determinant}\:\mathrm{is} \\ $$$$−{z}^{\mathrm{3}} \\ $$$$\:\:+\left(\mathrm{3}\beta−\mathrm{2}{p}\right){z}^{\mathrm{2}} \\ $$$$\:\:−\left({p}\alpha^{\mathrm{2}} +\mathrm{3}{q}\alpha+\mathrm{3}\beta^{\mathrm{2}} −\mathrm{4}{p}\beta+{p}^{\mathrm{2}} \right){z} \\ $$$$\:\:−\left({q}\alpha^{\mathrm{3}} −{p}\alpha^{\mathrm{2}} \beta−\mathrm{3}{q}\alpha\beta+{pq}\alpha−\beta^{\mathrm{3}} +\mathrm{2}{p}\beta^{\mathrm{2}} −{p}^{\mathrm{2}} \beta−{q}^{\mathrm{2}} \right){p} \\ $$$$\mathrm{we}\:\mathrm{want}\:\mathrm{the}\:\mathrm{square}\:\mathrm{and}\:\mathrm{the}\:\mathrm{linear}\:\mathrm{terms} \\ $$$$\mathrm{to}\:\mathrm{disappear}\:\mathrm{so}\:\mathrm{we}\:\mathrm{set}\:\mathrm{their}\:\mathrm{constants}\:\mathrm{zero} \\ $$$$\mathrm{to}\:\mathrm{get}\:\alpha\:\mathrm{and}\:\beta \\ $$$$\Rightarrow\:\beta=\frac{\mathrm{2}{p}}{\mathrm{3}};\:\alpha=−\frac{\mathrm{3}{q}}{\mathrm{2}{p}}\pm\frac{\sqrt{\mathrm{12}{p}^{\mathrm{3}} +\mathrm{81}{q}^{\mathrm{2}} }}{\mathrm{6}{p}} \\ $$$$\mathrm{this}\:\mathrm{leads}\:\mathrm{to} \\ $$$${z}^{\mathrm{3}} =\frac{\mathrm{8}{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{\mathrm{27}{q}^{\mathrm{4}} }{\mathrm{2}{p}^{\mathrm{3}} }+\mathrm{4}{q}^{\mathrm{2}} \pm\frac{{q}\sqrt{\mathrm{3}\left(\mathrm{4}{p}^{\mathrm{3}} +\mathrm{27}{q}^{\mathrm{2}} \right)^{\mathrm{3}} }}{\mathrm{18}{p}^{\mathrm{3}} } \\ $$$$\mathrm{Tschirnhaus}\:\mathrm{thought}\:\mathrm{he}\:\mathrm{could}\:\mathrm{solve}\:\mathrm{polynomes} \\ $$$$\mathrm{of}\:\mathrm{any}\:\mathrm{degree}\:\mathrm{with}\:\mathrm{this}\:\mathrm{method}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{getting} \\ $$$$\mathrm{harder}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{because}\:\mathrm{you}\:\mathrm{need}\:\mathrm{a}\:\mathrm{cubic} \\ $$$$\mathrm{substitution}\:\mathrm{to}\:\mathrm{eliminate}\:\mathrm{3}\:\mathrm{constants}\:\mathrm{and} \\ $$$$\mathrm{so}\:\mathrm{on}… \\ $$
Commented by maxmathsup by imad last updated on 10/Feb/19
$${thanks}\:{sir}\:. \\ $$
Commented by Tawa1 last updated on 10/Feb/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$