If x 2 + x + 1 = 0 , then the value of ( x + 1 x ) 4 + ( x 2 + 1 x 2 ) 4 + ( x 3 + 1 x 3 ) 4 + . . . + ( x 25 + 1 x 25 ) 4 is [2026]

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Solution

Q.
If $x^{2} + x + 1 = 0$ , then the value of ${(x + \frac{1}{x})}^{4} + {(x^{2} + \frac{1}{x^{2}})}^{4} + {(x^{3} + \frac{1}{x^{3}})}^{4} + . . . + {(x^{25} + \frac{1}{x^{25}})}^{4}$ is [2026]

1 128

2 162

3 145

4 175

Ans.
(3)

$x^{2} + x + 1 = 0$

$\Rightarrow x = ω or ω^{2}$

$∴$ $α = ω, β = ω^{2}$

$= {(ω + ω^{2})}^{4} + {(ω^{2} + ω^{4})}^{4} + {(ω^{3} + ω^{6})}^{4} + \dots + {(ω^{25} + ω^{50})}^{4}$

$= [{(ω + ω^{2})}^{4} + {(ω^{2} + ω^{4})}^{4} + {(ω^{4} + ω^{8})}^{4} + \dots + {(ω^{25} + ω^{50})}^{4}]$ $+ [{(ω^{3} + ω^{6})}^{4} + {(ω^{6} + ω^{12})}^{4} + {(ω^{9} + ω^{18})}^{4} + \dots + {(ω^{24} + ω^{48})}^{4}]$

$= \underset{17 times}{\underset{⏟}{[1 + 1 + 1 + \dots + 1]}} + \underset{8 times}{\underset{⏟}{[{(1 + 1)}^{4} + {(1 + 1)}^{4} + \dots + {(1 + 1)}^{4}]}}$

$= 17 + 128$

$= 145$

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