If x = 1−21 - \sqrt{2}1−2, find the value of (x−1x)3\Big(x - \dfrac{1}{x}\Big)^3(x−x1)3.
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Given,
x = 1 - 2\sqrt{2}2
∴1x=11−2\therefore \dfrac{1}{x} = \dfrac{1}{1 - \sqrt{2}}∴x1=1−21
Rationalizing,
⇒1x=11−2×1+21+2⇒1+212−(2)2⇒1+21−2⇒1+2−1⇒−(1+2)⇒−1−2.\Rightarrow \dfrac{1}{x} = \dfrac{1}{1 - \sqrt{2}} \times \dfrac{1 + \sqrt{2}}{1 + \sqrt{2}} \\[1em] \Rightarrow \dfrac{1 + \sqrt{2}}{1^2 - (\sqrt{2})^2} \\[1em] \Rightarrow \dfrac{1 + \sqrt{2}}{1 - 2} \\[1em] \Rightarrow \dfrac{1 + \sqrt{2}}{-1} \\[1em] \Rightarrow -(1 + \sqrt{2}) \\[1em] \Rightarrow -1 - \sqrt{2}.⇒x1=1−21×1+21+2⇒12−(2)21+2⇒1−21+2⇒−11+2⇒−(1+2)⇒−1−2.
Substituting values we get :
⇒(x−1x)3=[1−2−(−1−2)]3=[1+1−2+2]3=[2]3=8.\Rightarrow \Big(x - \dfrac{1}{x}\Big)^3 = [1 - \sqrt{2} - (-1 - \sqrt{2})]^3 \\[1em] = [1 + 1 - \sqrt{2} + \sqrt{2}]^3 \\[1em] = [2]^3 \\[1em] = 8.⇒(x−x1)3=[1−2−(−1−2)]3=[1+1−2+2]3=[2]3=8.
Hence, (x−1x)3\Big(x - \dfrac{1}{x}\Big)^3(x−x1)3 = 8.
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If m = 13−22 and n=13+22\dfrac{1}{3 - 2\sqrt{2}} \text{ and } n = \dfrac{1}{3 + 2\sqrt{2}}3−221 and n=3+221, find :
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(ii) n2
(iii) mn
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(i) 1x\dfrac{1}{x}x1
(ii) x+1xx + \dfrac{1}{x}x+x1
(iii) (x+1x)2\Big(x + \dfrac{1}{x}\Big)^2(x+x1)2
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(ii) 13+22\dfrac{1}{3 + 2\sqrt{2}}3+221
(iii) 2−33\dfrac{2 - \sqrt{3}}{\sqrt{3}}32−3
