Can anyone find any error in my attempt to improve upon the Cardano′s cubic formula.. or as an alternate to the trigonometric solution... here it goes: X^3 −pX−q=0 let (X/( (√p))) = x ; and with (q/(p(√p))) = c , x^3 −x−c=0 let x=(k+1)(p+q) =(p+kq) + (kp+q) p^3 +k^3 q^3 +3kpq(p+kq)+ k^3 p^3 +q^3 +3kpq(kp+q) −(p+kq)−(kp+q)=c ⇒ (1+k^3 )(p^3 +q^3 )+ (p+kq)(3kpq−1)+ (kp+q)(3kpq−1)= c let 3kpq=1 ⇒ p^3 +q^3 =(c/(1+k^3 )) & p^3 q^3 =(1/(27k^3 )) ; hence p^3 , q^3 = (c/(2(1+k^3 )))±(√((c^2 /(4(1+k^3 )^2 ))−(1/(27k^3 )))) lets choose upon a value of k such that D=0 ⇒ 4(1+k^3 )^2 =27c^2 k^3 .....(I) (just a quadratic..) first for c^2 >(8/(27)) we always can get two real k values, and even p=q then. x=(k+1)(p+q) = 2(k+1)p x=2(k+1)[(c/(2(1+k^3 )))]^(1/3) but simply pq=(1/(3k)) hence x=2(k+1)p = ((2(k+1))/( (√(3k)))) . ________________________ even for c=1 i dint get a correct answer, please help error-freeing it. (Thanks!)


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Question Number 128007 by ajfour last updated on 03/Jan/21

$C a n a n y o n e f i n d a n y e r r o r i n$ $m y a t t e m p t t o i m p r o v e u p o n$ $t h e {C a r d a n o}^{'} s c u b i c f o r m u l a . .$ $o r a s a n a l t e r n a t e t o t h e$ $t r i g o n o m e t r i c s o l u t i o n . . .$ $h e r e i t g o e s :$ $X^{3} - p X - q = 0$ $l e t \frac{X}{\sqrt{p}} = x; a n d w i t h \frac{q}{p \sqrt{p}} = c,$ $x^{3} - x - c = 0$ $l e t x = (k + 1) (p + q)$ $= (p + k q) + (k p + q)$ $p^{3} + k^{3} q^{3} + 3 k p q (p + k q) +$ $k^{3} p^{3} + q^{3} + 3 k p q (k p + q)$ $- (p + k q) - (k p + q) = c$ $\Rightarrow$ $(1 + k^{3}) (p^{3} + q^{3}) +$ $(p + k q) (3 k p q - 1) +$ $(k p + q) (3 k p q - 1) = c$ $l e t 3 k p q = 1 \Rightarrow$ $p^{3} + q^{3} = \frac{c}{1 + k^{3}}$ $& p^{3} q^{3} = \frac{1}{27 k^{3}}; h e n c e$ $p^{3}, q^{3} =$ $\frac{c}{2 (1 + k^{3})} \pm \sqrt{\frac{c^{2}}{4 {(1 + k^{3})}^{2}} - \frac{1}{27 k^{3}}}$ $l e t s c h o o s e u p o n a v a l u e o f k$ $s u c h t h a t D = 0$ $\Rightarrow 4 {(1 + k^{3})}^{2} = 27 c^{2} k^{3} . . . . . (I)$ $(j u s t a q u a d r a t i c . .)$ $f i r s t f o r c^{2} > \frac{8}{27} w e a l w a y s c a n$ $g e t t w o r e a l k v a l u e s, a n d e v e n$ $p = q t h e n .$ $x = (k + 1) (p + q)$ $= 2 (k + 1) p$ $x = 2 (k + 1) {[\frac{c}{2 (1 + k^{3})}]}^{1 / 3}$ $b u t s i m p l y p q = \frac{1}{3 k} h e n c e$ $x = 2 (k + 1) p = \frac{2 (k + 1)}{\sqrt{3 k}} .$ $________________________$ $e v e n f o r c = 1 i d i n t g e t a$ $c o r r e c t a n s w e r, p l e a s e h e l p$ $e r r o r - f r e e i n g i t . (T h a n k s!)$
Commented by MJS_new last updated on 04/Jan/21

$after substituting x = (k + 1) (p + q) I {don}^{'} t get$ $your next equation .$ $p, q are given \Rightarrow from the moment you set$ $3 k p q = 1 also k is given \Rightarrow x is given$
Commented by ajfour last updated on 04/Jan/21

$b e c a u s e w e p u t 3 k p q = 1$ $a n d u s e t h e e q u a t i o n, s o w e g e t$ $p^{3} + q^{3} = \frac{c}{2 (1 + k^{3})}$ $\Rightarrow i f w e g i v e a v a l u e t o k t h e n$ $x = (1 + k) (p + q) i s o b t a i n e d$ $i n a c c o r d a n c e w i t h t h e e q .$ $x^{3} = x + c ◼$
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