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Mathematics

If x = 3 + 222\sqrt{2}, find :

(i) 1x\dfrac{1}{x}

(ii) x−1xx - \dfrac{1}{x}

(iii) (x−1x)3\Big(x - \dfrac{1}{x}\Big)^3

(iv) x3−1x3x^3 - \dfrac{1}{x^3}

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Answer

(i) Solving,

1x=13+22\dfrac{1}{x} = \dfrac{1}{3 + 2\sqrt{2}} \\[1em]

Rationalizing,

⇒13+22×3−223−22⇒3−2232−(22)2⇒3−229−8⇒3−22.\Rightarrow \dfrac{1}{3 + 2\sqrt{2}} \times \dfrac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} \\[1em] \Rightarrow \dfrac{3- 2\sqrt{2}}{3^2 - (2\sqrt{2})^2} \\[1em] \Rightarrow \dfrac{3 - 2\sqrt{2}}{9 - 8} \\[1em] \Rightarrow 3 - 2\sqrt{2}.

Hence, 1x=3−22\dfrac{1}{x} = 3 - 2\sqrt{2}.

(ii) Solving,

⇒x−1x=3+22−(3−22)=3−3+22+22=42.\Rightarrow x - \dfrac{1}{x} = 3 + 2\sqrt{2} - (3 - 2\sqrt{2}) \\[1em] = 3 - 3 + 2\sqrt{2} + 2\sqrt{2} \\[1em] = 4\sqrt{2}.

Hence, x−1x=42x - \dfrac{1}{x} = 4\sqrt{2}.

(iii) Solving,

⇒(x−1x)3=(42)3=1282.\Rightarrow \Big(x - \dfrac{1}{x}\Big)^3 = (4\sqrt{2})^3 \\[1em] = 128\sqrt{2}.

Hence, (x−1x)3=1282\Big(x - \dfrac{1}{x}\Big)^3 = 128\sqrt{2}.

(iv) By formula,

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

Substituting values we get :

⇒x3−1x3=(42)3+3×42=1282+122=1402.\Rightarrow x^3 - \dfrac{1}{x^3} = (4\sqrt{2})^3 + 3 \times 4\sqrt{2} \\[1em] = 128\sqrt{2} + 12\sqrt{2} \\[1em] = 140\sqrt{2}.

Hence, x3−1x3=1402x^3 - \dfrac{1}{x^3} = 140\sqrt{2}.

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