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Question Number 125577 by ajfour last updated on 12/Dec/20

Commented by ajfour last updated on 12/Dec/20

$I f t h e c u b i c {c u r v e}^{'} s e q . i s$ $y = x^{3} - x - c$ $a n d {p a r a b o l a}^{'} s e q . i s$ $y = {(x - h)}^{2} + (\frac{2}{3 \sqrt{3}} - c), t h e n$ $w i t h t h e h e l p o f a s u i t a b l e v a l u e s$ $o f h a n d s,$ $f i n d t h e p o s i t i v e r o o t, x = p$ $o f t h e c u b i c .$
	Answered by ajfour last updated on 12/Dec/20
	$y=x^3 −x−c y=(x−h)^2 +k p^3 =p+c ..(i) (s−h)^2 +k=s^3 −s−c let s=p+m ⇒ p^3 +3mp^2 +3m^2 p+m^3 −p−m−c−k −p^2 −2mp−m^2 +2hp+2mh−h^2 =0 let 3m=1_(−) ⇒ p(2h−(1/3))−(8/(27))−k−(h−(1/3))^2 =0 ⇒ p=(((h−(1/3))^2 +k+(8/(27)))/(2h−(1/3))) but p^3 =p+c ⇒ {(h−(1/3))^2 +k+(8/(27))}^3 ={(h−(1/3))^2 +k+(8/(27))}(2h−(1/3))^2 +c(2h−(1/3))^3 let k=((2h)/3) (h^2 +((11)/(27)))^3 =(h^2 +((11)/(27)))(2h−(1/3))^2 +c(2h−(1/3))^3 (h^2 +((11)/(27)))(h^4 +((22h^2 )/(27))+((121)/(729))−4h^2 +((4h)/3)−(1/9)) =c(2h−(1/3))^3 .....$
	$y = x^{3} - x - c$ $y = {(x - h)}^{2} + k$ $p^{3} = p + c . . (i)$ ${(s - h)}^{2} + k = s^{3} - s - c$ $l e t s = p + m \Rightarrow$ $p^{3} + 3 {m p}^{2} + 3 m^{2} p + m^{3} - p - m - c - k$ $- p^{2} - 2 m p - m^{2} + 2 h p + 2 m h - h^{2} = 0$ $\underline{l e t 3 m = 1} \Rightarrow$ $p (2 h - \frac{1}{3}) - \frac{8}{27} - k - {(h - \frac{1}{3})}^{2} = 0$ $\Rightarrow p = \frac{{(h - \frac{1}{3})}^{2} + k + \frac{8}{27}}{2 h - \frac{1}{3}}$ $b u t p^{3} = p + c \Rightarrow$ ${{(h - \frac{1}{3})}^{2} + k + \frac{8}{27}}^{3}$ $= {{(h - \frac{1}{3})}^{2} + k + \frac{8}{27}} {(2 h - \frac{1}{3})}^{2}$ $+ c {(2 h - \frac{1}{3})}^{3}$ $l e t k = \frac{2 h}{3}$ ${(h^{2} + \frac{11}{27})}^{3} = (h^{2} + \frac{11}{27}) {(2 h - \frac{1}{3})}^{2}$ $+ c {(2 h - \frac{1}{3})}^{3}$ $(h^{2} + \frac{11}{27}) (h^{4} + \frac{22 h^{2}}{27} + \frac{121}{729} - 4 h^{2} + \frac{4 h}{3} - \frac{1}{9})$ $= c {(2 h - \frac{1}{3})}^{3}$ $. . . . .$
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