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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
If $x^2 + x + 1 = 0$, then the value of
$\left(x + \frac{1}{x}\right)^4 + \left(x^2 + \frac{1}{x^2}\right)^4 + \left(x^3 + \frac{1}{x^3}\right)^4 + \cdots + \left(x^{25} + \frac{1}{x^{25}}\right)^4$
is:
Solution
$x^2 + x + 1 = 0$
$\Rightarrow x = \omega$ or $\omega^2$
$\therefore \alpha = \omega,; \beta = \omega^2$
$= (\omega + \omega^2)^4 + (\omega^2 + \omega)^4 + (\omega^4 + \omega^2)^4 + \cdots$
$= [1 + 1 + \cdots + 1] + [(1+1)^4 + (-1+1)^4 + \cdots]$
$= 17 + 128 = 145$
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