x-2-4x-1-0-x-3-1-x-3-

Question Number 151344 by mathdanisur last updated on 20/Aug/21

x^{2} - 4 x + 1 = 0 \Rightarrow x^{3} + \frac{1}{x^{3}} = ?

Answered by dumitrel last updated on 20/Aug/21

x - 4 + \frac{1}{x} = 0 \Rightarrow x + \frac{1}{x} = 4 \Rightarrow

64 = x^{3} + 3 x \cdot \frac{1}{x} (x + \frac{1}{x}) + \frac{1}{x^{3}} \Rightarrow

64 = x^{3} + 12 + \frac{1}{x^{3}} \Rightarrow x^{3} + \frac{1}{x^{3}} = 52

Commented by mathdanisur last updated on 20/Aug/21

Thank you S er cool

Answered by Rasheed.Sindhi last updated on 20/Aug/21

x^{2} + 1 = 4 x

{(x^{2} + 1 = 4 x)}^{3}

x^{6} + 1 + {3 x}^{2} (x^{2} + 1) = {64 x}^{3}

x^{6} + 1 + {3 x}^{2} (4 x) = {64 x}^{3}

x^{6} - {52 x}^{3} + 1 = 0

D i v i d i n g b y x^{3} t o b o t h s i d e s : (x \neq 0)

x^{3} - 52 + \frac{1}{x^{3}} = 0

x^{3} + \frac{1}{x^{3}} = 52

Commented by mathdanisur last updated on 20/Aug/21

Thank you S er