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Mathematics

If (x+1x)2=3\Big(x + \dfrac{1}{x}\Big)^2 = 3, show that (x3+1x3)=0\Big(x^3 + \dfrac{1}{x^3}\Big) = 0.

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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}.

We know that,

⇒(x3+1x3)=(x+1x)3−3(x+1x)\Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = \Big(x + \dfrac{1}{x}\Big)^3 - 3\Big(x + \dfrac{1}{x}\Big)

Case 1 :

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

Substituting values we get :

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

Case 2 :

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

Substituting values we get :

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

Hence, proved that (x3+1x3)=0\Big(x^3 + \dfrac{1}{x^3}\Big) = 0.

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