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Question Number 129832 by ajfour last updated on 19/Jan/21

Commented by ajfour last updated on 19/Jan/21

$I f p, q, r a r e r o o t s o f y = x^{3} - x - c$ $t h e n f i n d p o r q o r r, w i t h t h e$ $h e l p o f s .$
Commented by MJS_new last updated on 20/Jan/21

$my opinion :$ $x^{4} + {a x}^{2} + b x + c = 0$ $can be solved exactly if we can find exact$ $factors α, β, γ with$ $(x^{2} - α x - β) (x^{2} + α x - γ) = x^{4} + {a x}^{2} + b x + c$ $\Leftrightarrow$ $we must find at least one u s e a b l e exact$ $solution for$ $y^{3} + P y + Q = 0 with P = - \frac{a^{2} + 12 c}{3} \land Q = - \frac{2 a^{3} - 72 a c + 27 b^{2}}{27}$ $now what you seem to want is finding s in$ $order to solve the following :$ $(X^{3} - X - C) (X - s) = 0$ $\Leftrightarrow$ $X^{4} - {s X}^{3} - X^{2} - (C - s) X + C s = 0 \Leftrightarrow$ $[X = x + \frac{s}{4}]$ $\Leftrightarrow x^{4} - \frac{3 s^{2} + 8}{8} x^{2} - \frac{s^{3} - 4 s + 8 C}{8} x - \frac{3 s (s^{3} - 16 s - 64 C)}{256} = 0$ $\Rightarrow we must be able to find a solution of$ $y^{3} + P y + Q = 0$ $with$ $P = - \frac{3 s^{2} + 9 C s + 1}{3}$ $Q = - \frac{27 {C s}^{3} + 18 s^{2} + 27 C s + 37 C^{2} - 2}{27}$ $try with$ $C = \frac{1}{3}$ $I {don}^{'} t think you can find a fitting s following$ $this path w i t h o u t solving another 3^{rd} degree$ $polynome$
	Answered by ajfour last updated on 20/Jan/21

	$y = x^{4} - {s x}^{3} - x^{2} + (s - c) x + c s$ $\frac{d y}{d x} = 4 x^{3} - 3 {s x}^{2} - 2 x + s - c$ $\frac{d^{2} y}{{d x}^{2}} = 12 x^{2} - 6 s x - 2$ $l e t 4 x^{3} - 3 {s x}^{2} - 2 x + s - c$ $= s - c$ $\Rightarrow 4 x^{2} - 3 s x - 2 = 0$ $3 {s x}^{2} - 2 x - 4 c = 0$ $\Rightarrow (8 c - 3 s) x = 6 c s - 2$ $\Rightarrow x = \frac{6 c s - 2}{8 c - 3 s}$ $4 {(6 c s - 2)}^{2} - 3 s (6 c s - 2) (8 c - 3 s)$ $- 2 {(8 c - 3 s)}^{2} = 0$ $. . .$
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