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Question Number 80881 by ajfour last updated on 07/Feb/20

$v = - \frac{(2 b + 3 c p) p}{3 + {b p}^{2}}$ $\frac{3 v^{2} + 2 b p v + 3 c p + b}{1 + {b p}^{2} + {c p}^{3}} > 0$ $b, c \in R, b < 0$ $A n y n o n - z e r o r e a l v a l u e o f p$ $i n t e r m s o f b, c o b e y i n g a b o v e$ $c o n d i t i o n ?$
	Answered by ajfour last updated on 07/Feb/20
	$let p=−(b/c) v=((−b^2 )/(c(3+(b^3 /c^2 ))))=−((b^2 c)/(3c^2 +b^3 )) S=(((b^2 c)/(3c^2 +b^3 )))^2 +((2b^2 )/c)(((b^2 c)/(3c^2 +b^3 )))−2b let me test it x^3 −49x+120=0 let x=((u+v)/(1+pu)) u^3 +Su+((v^3 +bv+c)/(1+bp^2 +cp^3 ))=0 ⇒ u^3 +Su+T=0 D= (T^( 2) /4)+(S^3 /(27)) u={−(T/2)+(√D)}^(1/3) +{−(T/2)−(√D)}^(1/3) ........$
	$l e t p = - \frac{b}{c}$ $v = \frac{- b^{2}}{c (3 + \frac{b^{3}}{c^{2}})} = - \frac{b^{2} c}{3 c^{2} + b^{3}}$ $S = {(\frac{b^{2} c}{3 c^{2} + b^{3}})}^{2} + \frac{2 b^{2}}{c} (\frac{b^{2} c}{3 c^{2} + b^{3}}) - 2 b$ $l e t m e t e s t i t$ $x^{3} - 49 x + 120 = 0$ $l e t x = \frac{u + v}{1 + p u}$ $u^{3} + S u + \frac{v^{3} + b v + c}{1 + {b p}^{2} + {c p}^{3}} = 0$ $\Rightarrow u^{3} + S u + T = 0$ $D = \frac{T^{2}}{4} + \frac{S^{3}}{27}$ $u = {- \frac{T}{2} + \sqrt{D}}^{1 / 3} + {- \frac{T}{2} - \sqrt{D}}^{1 / 3}$ $. . . . . . . .$
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