x^3 +ax^2 +bx+c=0 Transform to t^3 +k=0 .


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Question Number 56467 by ajfour last updated on 17/Mar/19

$x^{3} + {a x}^{2} + b x + c = 0$ $T r a n s f o r m t o$ $t^{3} + k = 0 .$
	Answered by ajfour last updated on 17/Mar/19

	$l e t x = \frac{p t + q}{t + 1}$ $\Rightarrow {(p t + q)}^{3} + a {(p t + q)}^{2} (t + 1)$ $+ b (p t + q) {(t + 1)}^{2} + c {(t + 1)}^{3} = 0$ $(p^{3} + {a p}^{2} + b p + c) t^{3} + (3 p^{2} q + {a p}^{2} + 2 a p q + b q + 2 b p + 3 c) t^{2}$ $+ (3 {p q}^{2} + 2 a p q + {a q}^{2} + b p + 2 b q + 3 c) t$ $+ q^{3} + {a q}^{2} + b q + c = 0$ $L e t 3 p^{2} q + {a p}^{2} + 2 a p q + b q + 2 b p + 3 c = 0$ $& 3 {p q}^{2} + 2 a p q + {a q}^{2} + b p + 2 b q + 3 c = 0$ $s u b t r a c t i n g$ $3 p q + a (p + q) - b + 2 b = 0$ $\Rightarrow 3 p q + a (p + q) + b = 0 (i)$ $u s i n g t h i s$ $- p [a (p + q) + b] + {a p}^{2} + 2 a p q + b q$ $+ 2 b p + 3 c = 0$ $\Rightarrow a p q + b q + b p + 3 c = 0$ $\Rightarrow a p q + b (p + q) + 3 c = 0 (i i)$ $\Rightarrow p + q = \frac{\| \begin{matrix} 3 & - b \\ a & - 3 c \end{matrix} \|}{\| \begin{matrix} 3 & a \\ a & b \end{matrix} \|} = \frac{- 9 c + a b}{3 b - a^{2}} = s$ $\Rightarrow p q = \frac{\| \begin{matrix} - b & a \\ - 3 c & b \end{matrix} \|}{\| \begin{matrix} 3 & a \\ a & b \end{matrix} \|} = \frac{- b^{2} + 3 a c}{3 b - a^{2}} = m$ $\Rightarrow p, q a r e r o o t s o f e q .$ $z^{2} - s z + m = 0$ $z = \frac{s}{2} \pm \sqrt{\frac{s^{2}}{4} - m}$ $S o$ $(p^{3} + {a p}^{2} + b p + c) t^{3} + q^{3} + {a q}^{2} + b q + c = 0$ $t = - {(\frac{q^{3} + {a q}^{2} + b q + c}{p^{3} + {a p}^{2} + b p + c})}^{1 / 3}$ $x = \frac{p t + q}{t + 1} .$
	Commented by behi83417@gmail.com last updated on 17/Mar/19
	$let try for: [x^3 +2x^2 +3x+4=0] {a=2,b=3,c=4} p+q=((−9c+ab)/(3b−a^2 ))=((−9×4+2×3)/(9−4))=−6 pq=((−b^2 +3ac)/(3b−a^2 ))=((−9+3×2×4)/5)=+3 ⇒z^2 +6z+3=0⇒z=((−6±(√(36−12)))/2)= ⇒z=−3±(√6)⇒ { ((p=−3+(√6)=−0.55,q=−3−(√6)=−5.44)),((p=−3−(√6),q=−3+(√6))) :} t^3 =−((q^3 +aq^2 +bq+c)/(p^3 +ap^2 +bp+c))=−(((−5.44)^3 +2(−5.44)^2 +3(−5.44)+4)/((−0.55)^3 +2(−0.55)^2 +3(−0.55)+4))=40.924 t=(40.925)^(1/3) =3.446 x=((pt+q)/(t+1))=((−0.55×3.446−5.44)/(3.446+1))=−1.65[ok!]$
	$l e t t r y f o r : [x^{3} + 2 x^{2} + 3 x + 4 = 0]$ ${a = 2, b = 3, c = 4}$ $p + q = \frac{- 9 c + a b}{3 b - a^{2}} = \frac{- 9 \times 4 + 2 \times 3}{9 - 4} = - 6$ $p q = \frac{- b^{2} + 3 a c}{3 b - a^{2}} = \frac{- 9 + 3 \times 2 \times 4}{5} = + 3$ $\Rightarrow z^{2} + 6 z + 3 = 0 \Rightarrow z = \frac{- 6 \pm \sqrt{36 - 12}}{2} =$ $\Rightarrow z = - 3 \pm \sqrt{6} \Rightarrow {\begin{cases} p = - 3 + \sqrt{6} = - 0.55, q = - 3 - \sqrt{6} = - 5.44 \\ p = - 3 - \sqrt{6}, q = - 3 + \sqrt{6} \end{cases}$ $t^{3} = - \frac{q^{3} + {a q}^{2} + b q + c}{p^{3} + {a p}^{2} + b p + c} = - \frac{{(- 5.44)}^{3} + 2 {(- 5.44)}^{2} + 3 (- 5.44) + 4}{{(- 0.55)}^{3} + 2 {(- 0.55)}^{2} + 3 (- 0.55) + 4} = 40.924$ $t = {(40.925)}^{\frac{1}{3}} = 3.446$ $x = \frac{p t + q}{t + 1} = \frac{- 0.55 \times 3.446 - 5.44}{3.446 + 1} = - 1.65 [o k!]$
	Commented by mr W last updated on 17/Mar/19
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	Commented by behi83417@gmail.com last updated on 17/Mar/19

	Commented by behi83417@gmail.com last updated on 17/Mar/19

	$c o n g r a t u l a t i o n s s i r A j f o u r!$ $t h i s i s a p e r f e c t w a y f o r 3 r d d e g . e q$ $t h a n k s f o r s h a r i n g t h i s k n o w l e d g e s .$
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