If ${(\frac{x + 1}{x})}^{3} = 3$ then find the value of $x^{206} + x^{200} + x^{90} + x^{84} + x^{18} + x^{12} + x^{6} + 1$

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Solution

${(x + \frac{1}{x})}^{2} = 3 x^{2} + 2 + \frac{1}{x^{2}} = 3 x^{2} - 1 + \frac{1}{x^{2}} = 0 x^{4} - x^{2} + 1 = 0$
Or, $x^{206} + x^{200} + x^{90} + x^{84} + x^{18} + x^{12} + x^{6} + 1$
Or, $x^{200} + (x^{200} (x^{6} + 1) + x^{84} (x^{6} + 1) + x^{12} + (x^{6} + 1) + (x^{6} + 1) =$
Or, $(x^{200} + x^{84} + x^{12} + 1) (x^{6} + 1) =$
Or, $(x^{200} + x^{84} + x^{12} + 1) (x^{2} + 1) (x^{4} - x^{2} + 1) =$
Or, $0$

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Similar questions

{(x + \frac{1}{x})}^{2} = 3

x^{206} + x^{200} + x^{90} + x^{84} + x^{18} + x^{12} + x^{6} + 1

x + \frac{1}{x} = \sqrt{3}

x^{18} + x^{12} + x^{6} + 1

{(x + \frac{1}{x})}^{2} = 3

x^{18} + x^{12} + x^{6} + 1

x^{2} + 1 = 0

x^{18} + x^{12} + x^{6} + 1

Find the coefficient of $x^{90}$ in $(x + x^{2} + x^{3} + . . . . . . . . . .)^{2} (x^{5} + x^{6} + . . . . . . . . . . .)^{9}$

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