x^3 +ax^2 +bx+c=0 Let x=((pt+q)/(t+1)) p^3 t^3 +3p^2 qt^2 +3pq^2 t+q^3 +a(t+1)(p^2 t^2 +2pqt+q^2 ) +b(pt+q)(t^2 +2t+1) +c(t^3 +3t^2 +3t+1) = 0 ⇒ (p^3 +ap^2 +bp+c)t^3 +(3p^2 q+ap^2 +2apq+bq+2bp+3c)t^2 +(3q^2 p+aq^2 +2apq+bp+2bq+3c)t +(q^3 +aq^2 +bq+c) = 0 Let coeffs. of t^2 and t be zero. Subtracting and adding them 3pq+a(p+q)+b=0 & 3pq(p+q)+a{(p+q)^2 −2pq} +4apq+3b(p+q)+6c = 0 lets call pq=m , p+q=s ⇒ 3m+as+b=0 ....(i) 3ms+a(s^2 −2m)+4am+3bs+6c=0 ⇒ am+bs+3c=0 ....(ii) ⇒ s=((9c−ab)/(a^2 −3b)) ; m=((b^2 −3ac)/(a^2 −3b)) Now p,q are roots of eq. z^2 −sz+m=0 p,q = (s/2)±(√((s^2 /4)−m)) t^3 =−(((q^3 +aq^2 +bq+c)/(p^3 +ap^2 +bq+c))) (t≠−1) x=((pt+q)/(t+1)) .


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Question Number 74339 by ajfour last updated on 22/Nov/19
$x^3 +ax^2 +bx+c=0 Let x=((pt+q)/(t+1)) p^3 t^3 +3p^2 qt^2 +3pq^2 t+q^3 +a(t+1)(p^2 t^2 +2pqt+q^2 ) +b(pt+q)(t^2 +2t+1) +c(t^3 +3t^2 +3t+1) = 0 ⇒ (p^3 +ap^2 +bp+c)t^3 +(3p^2 q+ap^2 +2apq+bq+2bp+3c)t^2 +(3q^2 p+aq^2 +2apq+bp+2bq+3c)t +(q^3 +aq^2 +bq+c) = 0 Let coeffs. of t^2 and t be zero. Subtracting and adding them 3pq+a(p+q)+b=0 & 3pq(p+q)+a{(p+q)^2 −2pq} +4apq+3b(p+q)+6c = 0 lets call pq=m , p+q=s ⇒ 3m+as+b=0 ....(i) 3ms+a(s^2 −2m)+4am+3bs+6c=0 ⇒ am+bs+3c=0 ....(ii) ⇒ s=((9c−ab)/(a^2 −3b)) ; m=((b^2 −3ac)/(a^2 −3b)) Now p,q are roots of eq. z^2 −sz+m=0 p,q = (s/2)±(√((s^2 /4)−m)) t^3 =−(((q^3 +aq^2 +bq+c)/(p^3 +ap^2 +bq+c))) (t≠−1) x=((pt+q)/(t+1)) .$
$${x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${Let}\:\:\:{x}=\frac{{pt}+{q}}{{t}+\mathrm{1}} \\ $$$${p}^{\mathrm{3}} {t}^{\mathrm{3}} +\mathrm{3}{p}^{\mathrm{2}} {qt}^{\mathrm{2}} +\mathrm{3}{pq}^{\mathrm{2}} {t}+{q}^{\mathrm{3}} \\ $$$$+{a}\left({t}+\mathrm{1}\right)\left({p}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{2}{pqt}+{q}^{\mathrm{2}} \right) \\ $$$$+{b}\left({pt}+{q}\right)\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{1}\right) \\ $$$$+{c}\left({t}^{\mathrm{3}} +\mathrm{3}{t}^{\mathrm{2}} +\mathrm{3}{t}+\mathrm{1}\right)\:\:\:\:=\:\:\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\:\:\left({p}^{\mathrm{3}} +{ap}^{\mathrm{2}} +{bp}+{c}\right){t}^{\mathrm{3}} \\ $$$$+\left(\mathrm{3}{p}^{\mathrm{2}} {q}+{ap}^{\mathrm{2}} +\mathrm{2}{apq}+{bq}+\mathrm{2}{bp}+\mathrm{3}{c}\right){t}^{\mathrm{2}} \\ $$$$+\left(\mathrm{3}{q}^{\mathrm{2}} {p}+{aq}^{\mathrm{2}} +\mathrm{2}{apq}+{bp}+\mathrm{2}{bq}+\mathrm{3}{c}\right){t} \\ $$$$+\left({q}^{\mathrm{3}} +{aq}^{\mathrm{2}} +{bq}+{c}\right)\:=\:\mathrm{0} \\ $$$${Let}\:{coeffs}.\:{of}\:{t}^{\mathrm{2}} \:{and}\:{t}\:{be}\:{zero}. \\ $$$${Subtracting}\:{and}\:{adding}\:{them} \\ $$$$\:\mathrm{3}{pq}+{a}\left({p}+{q}\right)+{b}=\mathrm{0}\:\:\:\& \\ $$$$\mathrm{3}{pq}\left({p}+{q}\right)+{a}\left\{\left({p}+{q}\right)^{\mathrm{2}} −\mathrm{2}{pq}\right\} \\ $$$$\:+\mathrm{4}{apq}+\mathrm{3}{b}\left({p}+{q}\right)+\mathrm{6}{c}\:=\:\mathrm{0} \\ $$$${lets}\:{call}\:\:{pq}={m}\:,\:\:{p}+{q}={s}\:\:\Rightarrow \\ $$$$\:\:\mathrm{3}{m}+{as}+{b}=\mathrm{0}\:\:\:\:....\left({i}\right) \\ $$$$\mathrm{3}{ms}+{a}\left({s}^{\mathrm{2}} −\mathrm{2}{m}\right)+\mathrm{4}{am}+\mathrm{3}{bs}+\mathrm{6}{c}=\mathrm{0} \\ $$$$\Rightarrow\:{am}+{bs}+\mathrm{3}{c}=\mathrm{0}\:\:\:\:....\left({ii}\right) \\ $$$$\Rightarrow\:\:{s}=\frac{\mathrm{9}{c}−{ab}}{{a}^{\mathrm{2}} −\mathrm{3}{b}}\:\:\:;\:\:{m}=\frac{{b}^{\mathrm{2}} −\mathrm{3}{ac}}{{a}^{\mathrm{2}} −\mathrm{3}{b}} \\ $$$$\:{Now}\:\:{p},{q}\:\:{are}\:{roots}\:{of}\:{eq}. \\ $$$$\:\:\:{z}^{\mathrm{2}} −{sz}+{m}=\mathrm{0} \\ $$$$\:\:\:{p},{q}\:=\:\frac{{s}}{\mathrm{2}}\pm\sqrt{\frac{{s}^{\mathrm{2}} }{\mathrm{4}}−{m}} \\ $$$$\:\:{t}^{\mathrm{3}} =−\left(\frac{{q}^{\mathrm{3}} +{aq}^{\mathrm{2}} +{bq}+{c}}{{p}^{\mathrm{3}} +{ap}^{\mathrm{2}} +{bq}+{c}}\right)\:\:\:\:\:\left({t}\neq−\mathrm{1}\right) \\ $$$$\:\:\:{x}=\frac{{pt}+{q}}{{t}+\mathrm{1}}\:. \\ $$
Commented by ajfour last updated on 23/Nov/19
$considering x^3 −6x^2 +3x+10=0 a=−6, b=3, c=10 s=((90+18)/(27))=4 , m=((9+180)/(27))=7 p,q=2±i(√3) ; let p=2−i(√3) let t^3 =−(((8+12i(√3)−18−3(√3)i−6(1+4i(√3))+3(2+i(√3))+10)/(8−12i(√3)−18+3(√3)i−6(1−4i(√3))+3(2−i(√3))+10))) ⇒ t^3 = −(((−12(√3)i)/(12(√3)i))) ⇒ t=1,ω,ω^2 x_1 =(4/2) = 2 x_2 =(((2−i(√3))(−1+i(√3))/2+(2+i(√3)))/((1+i(√3))/2)) = (((1+3(√3)i)+(4+2(√3)i))/(1+i(√3))) = 5 x_3 =(((2−i(√3))(−1−i(√3))/2+(2+i(√3)))/((1−i(√3))/2)) =((−1+i(√3))/( 1−i(√3))) = −1 hence roots are x∈{2,5,−1} (true indeed!)$
$${considering}\:\:\:{x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{10}=\mathrm{0} \\ $$$${a}=−\mathrm{6},\:{b}=\mathrm{3},\:{c}=\mathrm{10} \\ $$$${s}=\frac{\mathrm{90}+\mathrm{18}}{\mathrm{27}}=\mathrm{4}\:\:,\:{m}=\frac{\mathrm{9}+\mathrm{180}}{\mathrm{27}}=\mathrm{7} \\ $$$${p},{q}=\mathrm{2}\pm{i}\sqrt{\mathrm{3}}\:\:\:;\:\:{let}\:{p}=\mathrm{2}−{i}\sqrt{\mathrm{3}} \\ $$$${let}\:\:{t}^{\mathrm{3}} =−\left(\frac{\mathrm{8}+\mathrm{12}{i}\sqrt{\mathrm{3}}−\mathrm{18}−\mathrm{3}\sqrt{\mathrm{3}}{i}−\mathrm{6}\left(\mathrm{1}+\mathrm{4}{i}\sqrt{\mathrm{3}}\right)+\mathrm{3}\left(\mathrm{2}+{i}\sqrt{\mathrm{3}}\right)+\mathrm{10}}{\mathrm{8}−\mathrm{12}{i}\sqrt{\mathrm{3}}−\mathrm{18}+\mathrm{3}\sqrt{\mathrm{3}}{i}−\mathrm{6}\left(\mathrm{1}−\mathrm{4}{i}\sqrt{\mathrm{3}}\right)+\mathrm{3}\left(\mathrm{2}−{i}\sqrt{\mathrm{3}}\right)+\mathrm{10}}\right) \\ $$$$\Rightarrow\:{t}^{\mathrm{3}} =\:−\left(\frac{−\mathrm{12}\sqrt{\mathrm{3}}{i}}{\mathrm{12}\sqrt{\mathrm{3}}{i}}\right)\:\:\:\:\Rightarrow\:\:{t}=\mathrm{1},\omega,\omega^{\mathrm{2}} \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{4}}{\mathrm{2}}\:=\:\mathrm{2}\:\:\:\: \\ $$$${x}_{\mathrm{2}} =\frac{\left(\mathrm{2}−{i}\sqrt{\mathrm{3}}\right)\left(−\mathrm{1}+{i}\sqrt{\mathrm{3}}\right)/\mathrm{2}+\left(\mathrm{2}+{i}\sqrt{\mathrm{3}}\right)}{\left(\mathrm{1}+{i}\sqrt{\mathrm{3}}\right)/\mathrm{2}} \\ $$$$\:\:\:\:=\:\frac{\left(\mathrm{1}+\mathrm{3}\sqrt{\mathrm{3}}{i}\right)+\left(\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}{i}\right)}{\mathrm{1}+{i}\sqrt{\mathrm{3}}}\:=\:\mathrm{5} \\ $$$$\:{x}_{\mathrm{3}} =\frac{\left(\mathrm{2}−{i}\sqrt{\mathrm{3}}\right)\left(−\mathrm{1}−{i}\sqrt{\mathrm{3}}\right)/\mathrm{2}+\left(\mathrm{2}+{i}\sqrt{\mathrm{3}}\right)}{\left(\mathrm{1}−{i}\sqrt{\mathrm{3}}\right)/\mathrm{2}} \\ $$$$\:\:\:\:\:\:=\frac{−\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\:\:\:\:\mathrm{1}−{i}\sqrt{\mathrm{3}}}\:=\:−\mathrm{1}\: \\ $$$${hence}\:\:{roots}\:{are}\:\:\:{x}\in\left\{\mathrm{2},\mathrm{5},−\mathrm{1}\right\} \\ $$$$\left({true}\:{indeed}!\right)\:\:\:\:\:\:\: \\ $$
Commented by ajfour last updated on 23/Nov/19

$${considering}\:\:\:{x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{10}=\mathrm{0} \\ $$$${again}\:{and}\:{using}\:{Cardano}'{s}\:{method} \\ $$$$\:\:{let}\:\:{x}={t}+\mathrm{2}\:\:\:\Rightarrow \\ $$$$\:\:{t}^{\mathrm{3}} +\mathrm{12}{t}+\mathrm{8}−\mathrm{24}{t}−\mathrm{24}+\mathrm{3}{t}+\mathrm{6}+\mathrm{10}=\mathrm{0} \\ $$$$\Rightarrow\:\:{t}^{\mathrm{3}} −\mathrm{9}{t}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\:{t}=\pm\mathrm{3},\:\mathrm{0} \\ $$$$\Rightarrow\:\:{x}=\:−\mathrm{1},\:\mathrm{2},\:\mathrm{5}\:\:\:\:\left({easy}\:{enough},\right. \\ $$$$\left.\:\:\:\:\:\:{n}\:{quite}\:{disheartning}!\right) \\ $$
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