1.

If \(x + \frac{1}{x} = \sqrt 3\), then the value of x18 + x12 + x6 + 1 is:

Answer» Correct Answer - Option 1 : 0

Given:

\(x + \frac{1}{x} = \;\surd 3\)

Calculation:

Taking cube both side of the equation,

\({x^3} + \;\frac{1}{{{x^3}}}\; + \;3x × \frac{1}{x}\left( {x\; + \;\frac{1}{x}} \right) = 3\sqrt 3 \)

⇒ \({x^3} + \;\frac{1}{{{x^3}}} + 3\sqrt 3 = 3\surd 3\)

⇒ \(\frac{{{x^6} + 1}}{{{x^3}}}\; = \;0\)

⇒ x6 + 1 = 0

⇒ x6 = - 1

x18 + x12 + x6 + 1

⇒ x12(x6 + 1) + (x6 + 1)

⇒ x12 × 0 + 0

⇒ 0

∴ The value of x18 + x12 + x6 + 1 is 0.


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