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...


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Question Number 54775 by MJS last updated on 10/Feb/19

$found something interesting, it was published$ $by Tschirnhaus in 1683$ $we can reduce$ $x^{3} + {a x}^{2} + b x + c = 0$ $(1) to$ $y^{3} + p y + q = 0$ $(2) and further to$ $z^{3} = t$ $(1) is the well known linear substitution$ $y = x + \frac{a}{3} \to x = y - \frac{a}{3}$ $\Rightarrow y^{3} - \frac{a^{2} - 3 b}{3} y + \frac{2 a^{3} - 9 a b + 27 c}{27} = 0$ $p = - \frac{a^{2} - 3 b}{3} and q = \frac{2 a^{3} - 9 a b + 27 c}{27}$ $\Rightarrow y^{3} + p y + q = 0$ $(2) quadratic substitution$ $z = y^{2} + α y + β \to y^{2} + α y + (β - z) = 0$ $we could solve this for y and then plug in$ $above . . . (Tschirnhaus did) but {there}^{'} s an$ $easier way : we calculate the determinant of$ $the Sylvester Matrix$ $we have$ $(a) 1 y^{3} + 0 y^{2} + p y + q = 0$ $(b) 0 y^{3} + y^{2} + α y + (β - z) = 0$ $the matrix is$ $[\begin{matrix} 1 & 0 & p & q & 0 \\ 0 & 1 & 0 & p & q \\ 0 & 1 & α & β - z & 0 \\ 0 & 0 & 1 & α & β - z \\ 1 & α & β - z & 0 & 0 \end{matrix}]$ $the determinant is$ $- z^{3}$ $+ (3 β - 2 p) z^{2}$ $- (p α^{2} + 3 q α + 3 β^{2} - 4 p β + p^{2}) z$ $- (q α^{3} - p α^{2} β - 3 q α β + p q α - β^{3} + 2 p β^{2} - p^{2} β - q^{2}) p$ $we want the square and the linear terms$ $to disappear so we set their constants zero$ $to get α and β$ $\Rightarrow β = \frac{2 p}{3}; α = - \frac{3 q}{2 p} \pm \frac{\sqrt{12 p^{3} + 81 q^{2}}}{6 p}$ $this leads to$ $z^{3} = \frac{8 p^{3}}{27} + \frac{27 q^{4}}{2 p^{3}} + 4 q^{2} \pm \frac{q \sqrt{3 {(4 p^{3} + 27 q^{2})}^{3}}}{18 p^{3}}$ $Tschirnhaus thought he could solve polynomes$ $of any degree with this method but {it}^{'} s getting$ $harder to solve because you need a cubic$ $substitution to eliminate 3 constants and$ $so on . . .$
Commented by maxmathsup by imad last updated on 10/Feb/19

$t h a n k s s i r .$
Commented by Tawa1 last updated on 10/Feb/19

$God bless you sir$
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