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Mathematics

If (x+1x)2=3\Big(x + \dfrac{1}{x}\Big)^2 = 3, then x3+1x3x^3 + \dfrac{1}{x^3} =

  1. 9

  2. 3

  3. 2

  4. 0

Expansions

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Answer

Given,

⇒(x+1x)2=3⇒(x+1x)=±3\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 3 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big) = \pm \sqrt{3}

Case 1:

⇒(x+1x)=3\Rightarrow \Big(x + \dfrac{1}{x}\Big) = \sqrt{3}

Using identity,

⇒(x3+1x3)=(x+1x)3−3(x+1x)⇒(x3+1x3)=(3)3−3(3)⇒(x3+1x3)=(33)−3(3)⇒(x3+1x3)=0\Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = \Big(x + \dfrac{1}{x}\Big)^3 - 3\Big(x + \dfrac{1}{x}\Big) \\[1em] \Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = (\sqrt{3})^3 - 3(\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = (3\sqrt{3}) - 3(\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = 0

Case 2:

⇒(x+1x)=−3\Rightarrow \Big(x + \dfrac{1}{x}\Big) = -\sqrt{3}

Using identity,

⇒(x3+1x3)=(x+1x)3−3(x+1x)⇒(x3+1x3)=(−3)3−3(−3)⇒(x3+1x3)=(−33)+3(3)⇒(x3+1x3)=0\Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = \Big(x + \dfrac{1}{x}\Big)^3 - 3\Big(x + \dfrac{1}{x}\Big) \\[1em] \Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = (-\sqrt{3})^3 - 3(-\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = (-3\sqrt{3}) + 3(\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = 0

Hence, Option 4 is the correct option.

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