x-3-x-c-0-c-lt-2-3-3-Solve-by-a-method-other-than-trigonometric-solution-

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 \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 \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

\dots \dots

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