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Question Number 74339 by ajfour last updated on 22/Nov/19

x^{3} + {a x}^{2} + b x + c = 0

L e t x = \frac{p t + q}{t + 1}

p^{3} t^{3} + 3 p^{2} {q t}^{2} + 3 {p q}^{2} t + q^{3}

+ a (t + 1) (p^{2} t^{2} + 2 p q t + q^{2})

+ b (p t + q) (t^{2} + 2 t + 1)

+ c (t^{3} + 3 t^{2} + 3 t + 1) = 0

\Rightarrow

(p^{3} + {a p}^{2} + b p + c) t^{3}

+ (3 p^{2} q + {a p}^{2} + 2 a p q + b q + 2 b p + 3 c) t^{2}

+ (3 q^{2} p + {a q}^{2} + 2 a p q + b p + 2 b q + 3 c) t

+ (q^{3} + {a q}^{2} + b q + c) = 0

L e t c o e f f s . o f t^{2} a n d t b e z e r o .

S u b t r a c t i n g a n d a d d i n g t h e m

3 p q + a (p + q) + b = 0 &

3 p q (p + q) + a {{(p + q)}^{2} - 2 p q}

+ 4 a p q + 3 b (p + q) + 6 c = 0

l e t s c a l l p q = m, p + q = s \Rightarrow

3 m + a s + b = 0 \dots . (i)

3 m s + a (s^{2} - 2 m) + 4 a m + 3 b s + 6 c = 0

\Rightarrow a m + b s + 3 c = 0 \dots . (i i)

\Rightarrow s = \frac{9 c - a b}{a^{2} - 3 b}; m = \frac{b^{2} - 3 a c}{a^{2} - 3 b}

N o w p, q a r e r o o t s o f e q .

z^{2} - s z + m = 0

p, q = \frac{s}{2} \pm \sqrt{\frac{s^{2}}{4} - m}

t^{3} = - (\frac{q^{3} + {a q}^{2} + b q + c}{p^{3} + {a p}^{2} + b q + c}) (t \neq - 1)

x = \frac{p t + q}{t + 1} .

Commented by ajfour last updated on 23/Nov/19

c o n s i d e r i n g x^{3} - 6 x^{2} + 3 x + 10 = 0

a = - 6, b = 3, c = 10

s = \frac{90 + 18}{27} = 4, m = \frac{9 + 180}{27} = 7

p, q = 2 \pm i \sqrt{3}; l e t p = 2 - i \sqrt{3}

l e t t^{3} = - (\frac{8 + 12 i \sqrt{3} - 18 - 3 \sqrt{3} i - 6 (1 + 4 i \sqrt{3}) + 3 (2 + i \sqrt{3}) + 10}{8 - 12 i \sqrt{3} - 18 + 3 \sqrt{3} i - 6 (1 - 4 i \sqrt{3}) + 3 (2 - i \sqrt{3}) + 10})

\Rightarrow t^{3} = - (\frac{- 12 \sqrt{3} i}{12 \sqrt{3} i}) \Rightarrow t = 1, ω, ω^{2}

x_{1} = \frac{4}{2} = 2

x_{2} = \frac{(2 - i \sqrt{3}) (- 1 + i \sqrt{3}) / 2 + (2 + i \sqrt{3})}{(1 + i \sqrt{3}) / 2}

= \frac{(1 + 3 \sqrt{3} i) + (4 + 2 \sqrt{3} i)}{1 + i \sqrt{3}} = 5

x_{3} = \frac{(2 - i \sqrt{3}) (- 1 - i \sqrt{3}) / 2 + (2 + i \sqrt{3})}{(1 - i \sqrt{3}) / 2}

= \frac{- 1 + i \sqrt{3}}{1 - i \sqrt{3}} = - 1

h e n c e r o o t s a r e x \in {2, 5, - 1}

(t r u e i n d e e d!)

Commented by ajfour last updated on 23/Nov/19

c o n s i d e r i n g x^{3} - 6 x^{2} + 3 x + 10 = 0

a g a i n a n d u s i n g {C a r d a n o}^{'} s m e t h o d

l e t x = t + 2 \Rightarrow

t^{3} + 12 t + 8 - 24 t - 24 + 3 t + 6 + 10 = 0

\Rightarrow t^{3} - 9 t = 0

\Rightarrow t = \pm 3, 0

\Rightarrow x = - 1, 2, 5 (e a s y e n o u g h,

n q u i t e d i s h e a r t n i n g!)

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