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Special Integrals - 2

If x^2-5x+1=0...

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If x^2-5x+1=0 (x is not equal to 0),then the value of x^3+(1/x^3) is?

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x^{2} - 4 x + 1 = 0

,, then what is the value of

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

?

x^{2} - x + 1 = 0

, then the value of

{(x + \frac{1}{x})}^{2} + {(x^{2} + \frac{1}{x^{2}})}^{2} + {(x^{3} + \frac{1}{x^{3}})}^{2} + {(x^{5} + \frac{1}{x^{5}})}^{2}

x^{2} + x + 1 = 0

x^{3}

x_{1}, x_{2}, x_{3}

be the roots of the equation

x^{3} - x + 1 = 0

, then the value of

(\frac{1 + x_{1}}{1 - x_{1}}) (\frac{1 + x_{2}}{1 - x_{2}}) + (\frac{1 + x_{1}}{1 - x_{1}}) (\frac{1 + x_{3}}{1 - x_{3}}) + (\frac{1 + x_{2}}{1 - x_{2}}) (\frac{1 + x_{3}}{1 - x_{3}})

x^{2} + x + 1 = 0

, then the numerical value of

{(x + \frac{1}{x})}^{2} + {(x^{2} + \frac{1}{x^{2}})}^{2} + {(x^{3} + \frac{1}{x^{3}})}^{2} + {(x^{4} + \frac{1}{x^{4}})}^{2} + . . . . + {(x^{27} + \frac{1}{x^{27}})}^{2}

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