x^3 −x−c=0 ; [c<2/(3(√3))] (Solve by a method other than trigonometric solution).


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Question Number 126363 by ajfour last updated on 19/Dec/20

$x^{3} - x - c = 0; [c < 2 / (3 \sqrt{3})]$ $(S o l v e b y a m e t h o d o t h e r t h a n$ $t r i g o n o m e t r i c s o l u t i o n) .$
	Answered by mathmax by abdo last updated on 19/Dec/20
	$x^3 −x−c=0 let x=u+v so e⇒(u+v)^3 −(u+v)−c=0 ⇒ u^3 +v^3 +3uv(u+v)−(u+v)−c=0 ⇒ u^3 +v^3 −c +(3uv−1)(u+v)=0 ⇒ { ((u^3 +v^3 =c)),((3uv−1=0)) :} ⇒ { ((u^3 +v^3 =c ⇒ { ((u^3 +v^3 =c)),((u^3 v^3 =(1/(27)))) :})),((uv=(1/3))) :} so u^3 and v^3 are solution of t^2 −ct +(1/(27))=0 Δ=c^2 −(4/(27)) <0 due to c<(2/(3(√3))) ⇒t_1 =((c+i(√((4/(27))−c^2 )))/2) t_2 =((c−i(√((4/(27))−c^2 )))/2) ⇒x=^3 (√((c+i(√((4/(27))−c^2 )))/2)) +^3 (√((c−i(√((4/(27))−c^2 )))/2))$
	$x^{3} - x - c = 0 let x = u + v so e \Rightarrow {(u + v)}^{3} - (u + v) - c = 0 \Rightarrow$ $u^{3} + v^{3} + 3 uv (u + v) - (u + v) - c = 0 \Rightarrow$ $u^{3} + v^{3} - c + (3 uv - 1) (u + v) = 0 \Rightarrow {\begin{cases} u^{3} + v^{3} = c \\ 3 uv - 1 = 0 \end{cases}$ $\Rightarrow {\begin{cases} u^{3} + v^{3} = c \Rightarrow {\begin{cases} u^{3} + v^{3} = c \\ u^{3} v^{3} = \frac{1}{27} \end{cases} \\ uv = \frac{1}{3} \end{cases}$ $so u^{3} and v^{3} are solution of t^{2} - ct + \frac{1}{27} = 0$ $Δ = c^{2} - \frac{4}{27} < 0 due to c < \frac{2}{3 \sqrt{3}} \Rightarrow t_{1} = \frac{c + i \sqrt{\frac{4}{27} - c^{2}}}{2}$ $t_{2} = \frac{c - i \sqrt{\frac{4}{27} - c^{2}}}{2} \Rightarrow x =^{3} \sqrt{\frac{c + i \sqrt{\frac{4}{27} - c^{2}}}{2}} +^{3} \sqrt{\frac{c - i \sqrt{\frac{4}{27} - c^{2}}}{2}}$
	Commented by MJS_new last updated on 19/Dec/20

	${Cardano}^{'} s solution is not valid for D < 0$
	Commented by mathmax by abdo last updated on 20/Dec/20

	$why sir ?$
	Commented by MJS_new last updated on 20/Dec/20

	$x^{3} + p x + q = 0; D = \frac{p^{3}}{27} + \frac{q^{2}}{4}$ $if D < 0 we have 3 real roots we cannot find$ $using Cardano . see below article$
	Commented by MJS_new last updated on 20/Dec/20
	https://en.m.wikipedia.org/wiki/Casus_irreducibilis
	Commented by MJS_new last updated on 20/Dec/20

	${don}^{'} t get confused, in this article Δ is n o t$ $equal to the D I use . still {it}^{'} s true . you cannot$ $reduce the complex number to its real value$ $\Rightarrow the solution is possible to write down but$ $impossible to use$
	Commented by mr W last updated on 20/Dec/20

	${t h a t}^{'} s t h e p o i n t!$ $c e r t a i n l y w e c a n e x p r e s s t h e r o o t s$ $u s i n g {c a r d a n o}^{'} s f o r m u l a u s i n g$ $c o m p l e x n u m b e r s, b u t t o c a l c u l a t e$ $t h e r e a l v a l u e s o u t f r o m t h e f o r m u l a,$ $w e m u s t a t t h e e n d c o m e t o t h e$ $t r i g o n o m e t r i c e x p r e s s i o n f o r m .$ $t h e r e f o r e {c a r d a n o}^{'} s f o r m u l a i s n o t$ $r e a l l y u s e f u l i n t h i s c a s e .$
	Commented by mathmax by abdo last updated on 20/Dec/20

	$nevermind thank you sir$
	Commented by Tawa11 last updated on 23/Jul/21

	$great$
	Answered by ajfour last updated on 19/Dec/20
	$x=t+h ⇒ t^3 +3ht^2 +(3h^2 −1)t+h^3 −h−c=0 let t=p be also a root. ⇒ t^4 +(3h−p)t^3 +(3h^2 −1−3ph)t^2 +[h^3 −h−c−p(3h^2 −1)]t −p(h^3 −h−c)=0 lets add to it qt{t^3 +3ht^2 +(3h^2 −1)t+h^3 −h−c}=0 ⇒ (1+q)t^4 +(3h−p+3qh)t^3 + [3h^2 −1−3ph+q(3h^2 −1)]t^2 + [h^3 −h−c−p(3h^2 −1)+q(h^3 −h−c)]t −p(h^3 −h−c)=0 Now let 3ph=(1+q)(3h^2 −1);[coeff. of t^2 =0] ⇒ 3ht^4 +(6h^2 +1)t^3 + [3h(h^3 −h−c)−(3h^2 −1)^2 ]t −(3h^2 −1)(h^3 −h−c)=0 let h^3 +mh−c=0 ⇒ h^3 −h−c=−(m+1)h ⇒ 3ht^4 +(6h^2 +1)t^3 −[(3h^2 −1)^2 +3(m+1)h^2 ]t +(m+1)h(3h^2 −1)=0 ⇒ ......$
	$x = t + h \Rightarrow$ $t^{3} + 3 {h t}^{2} + (3 h^{2} - 1) t + h^{3} - h - c = 0$ $l e t t = p b e a l s o a r o o t . \Rightarrow$ $t^{4} + (3 h - p) t^{3} + (3 h^{2} - 1 - 3 p h) t^{2}$ $+ [h^{3} - h - c - p (3 h^{2} - 1)] t$ $- p (h^{3} - h - c) = 0$ $l e t s a d d t o i t$ $q t {t^{3} + 3 {h t}^{2} + (3 h^{2} - 1) t + h^{3} - h - c} = 0$ $\Rightarrow$ $(1 + q) t^{4} + (3 h - p + 3 q h) t^{3} +$ $[3 h^{2} - 1 - 3 p h + q (3 h^{2} - 1)] t^{2} +$ $[h^{3} - h - c - p (3 h^{2} - 1) + q (h^{3} - h - c)] t$ $- p (h^{3} - h - c) = 0$ $N o w l e t$ $3 p h = (1 + q) (3 h^{2} - 1); [c o e f f . o f t^{2} = 0]$ $\Rightarrow 3 {h t}^{4} + (6 h^{2} + 1) t^{3} +$ $[3 h (h^{3} - h - c) - {(3 h^{2} - 1)}^{2}] t$ $- (3 h^{2} - 1) (h^{3} - h - c) = 0$ $l e t h^{3} + m h - c = 0 \Rightarrow$ $h^{3} - h - c = - (m + 1) h$ $\Rightarrow 3 {h t}^{4} + (6 h^{2} + 1) t^{3}$ $- [{(3 h^{2} - 1)}^{2} + 3 (m + 1) h^{2}] t$ $+ (m + 1) h (3 h^{2} - 1) = 0 \Rightarrow$ $. . . . . .$
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