Question-180652

Question Number 180652 by Shrinava last updated on 14/Nov/22

Answered by aleks041103 last updated on 17/Nov/22

l e t d e t X = x \Rightarrow a = 1

d e t (A^{2}) = {(d e t A)}^{2} = 1 = d e t (B C) = d e t B d e t C

\Rightarrow b c = 1 \Rightarrow b = \frac{1}{c}

B^{2} = C A \Rightarrow b^{2} = c \Rightarrow \frac{1}{c^{2}} = c

C^{2} = A B \Rightarrow c^{2} = b \Rightarrow c^{2} = \frac{1}{c}

\Rightarrow c^{3} = 1 = b^{3}

\Rightarrow c^{2} = \frac{1}{c} = b, b^{3} = 1

\Rightarrow x^{4} + 6 b^{3} x^{3} - 14 c^{2} x^{2} + 6 x + 1 = 0

\Rightarrow x^{4} + 6 x^{3} - 14 {b x}^{2} + 6 x + 1 = 0

o b v . x \neq 0

\Rightarrow (x^{2} + \frac{1}{x^{2}}) + 6 (x + \frac{1}{x}) - 14 b = 0

y = x + \frac{1}{x}

y^{2} = x^{2} + \frac{1}{x^{2}} + 2

\Rightarrow y^{2} - 2 + 6 y - 14 b = 0

\Rightarrow y^{2} + 6 y - 2 (1 + 7 b) = 0

\Rightarrow x + \frac{1}{x} = \frac{- 6 \pm \sqrt{44 + 56 b}}{2} = - 3 \pm \sqrt{11 + 14 b}

\Rightarrow x^{2} - (- 3 \pm \sqrt{11 + 14 b}) x + 1 = 0

\Rightarrow x = \frac{1}{2} (- 3 \pm \sqrt{11 + 14 b} \pm \sqrt{{(- 3 \pm \sqrt{11 + 14 b})}^{2} - 4})

b^{3} = 1 \Rightarrow b = 1, e^{\pm \frac{2 i π}{3}}

f o r b = 1 :

x = \frac{1}{2} (- 3 \pm 5 \pm \sqrt{{(- 3 \pm 5)}^{2} - 4}) =

= 1, 4 \pm \sqrt{15}

\dots

Commented by Shrinava last updated on 17/Nov/22

Thank you very much dear professor

Your solutions are perfect