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Mathematics

If x=2−3x = 2 - \sqrt{3} then find the value of x3−1x3x^3 - \dfrac{1}{x^3}.

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Answer

Given,

⇒x=2−3∴1x=12−3⇒1x=12−3×2+32+3⇒1x=2+34−(3)2⇒1x=2+34−3=2+3.\Rightarrow x = 2 - \sqrt{3} \\[1em] \therefore \dfrac{1}{x} = \dfrac{1}{2 - \sqrt{3}} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{1}{2 - \sqrt{3}} \times \dfrac{2 + \sqrt{3}}{2 + \sqrt{3}} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{2 + \sqrt{3}}{4 - (\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{2 + \sqrt{3}}{4 - 3} = 2 + \sqrt{3}.

So,

x−1x=2−3−(2+3)=−23.x - \dfrac{1}{x} = 2 - \sqrt{3} - (2 + \sqrt{3}) = -2\sqrt{3}.

Cubing both sides we get,

⇒(x−1x)3=(−23)3⇒x3−1x3−3(x)(1x)(x−1x)=−243⇒x3−1x3−3×−23=−243⇒x3−1x3+63=−243⇒x3−1x3=−243−63⇒x3−1x3=−303.\Rightarrow \Big(x - \dfrac{1}{x}\Big)^3 = (-2\sqrt{3})^3 \\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} - 3(x)\Big(\dfrac{1}{x}\Big)\Big(x - \dfrac{1}{x}\Big) = -24\sqrt{3} \\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} - 3 \times -2\sqrt{3} = -24\sqrt{3} \\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} + 6\sqrt{3} = -24\sqrt{3} \\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} = -24\sqrt{3} - 6\sqrt{3} \\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} = -30\sqrt{3}.

Hence, x3−1x3=−303x^3 - \dfrac{1}{x^3} = -30\sqrt{3}.

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