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Mathematics

If x = 3−2\sqrt{3} - \sqrt{2}, find the value of :

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

(ii) x2+1x2x^2 + \dfrac{1}{x^2}

(iii) x3+1x3x^3 + \dfrac{1}{x^3}

(iv) x3+1x3−3(x2+1x2)+x+1xx^3 + \dfrac{1}{x^3} - 3\Big(x^2 + \dfrac{1}{x^2}\Big) + x + \dfrac{1}{x}

Rational Irrational Nos

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Answer

(i) Given,

x = 3−2\sqrt{3} - \sqrt{2}

1x=13−2\dfrac{1}{x} = \dfrac{1}{\sqrt{3} - \sqrt{2}}

Rationalizing,

⇒13−2×3+23+2⇒3+2(3)2−(2)2⇒3+23−2⇒3+2∴1x=3+2⇒x+1x=3−2+(3+2)=23.\Rightarrow \dfrac{1}{\sqrt{3} - \sqrt{2}} \times \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} \\[1em] \Rightarrow \dfrac{\sqrt{3} + \sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2} \\[1em] \Rightarrow \dfrac{\sqrt{3} + \sqrt{2}}{3 - 2} \\[1em] \Rightarrow \sqrt{3} + \sqrt{2} \\[1em] \therefore \dfrac{1}{x} = \sqrt{3} + \sqrt{2} \\[2em] \Rightarrow x + \dfrac{1}{x} = \sqrt{3} - \sqrt{2} + (\sqrt{3} + \sqrt{2}) \\[1em] = 2\sqrt{3}.

Hence, x+1x=23x + \dfrac{1}{x} = 2\sqrt{3}.

(ii) Solving,

⇒x2+1x2=(3−2)2+(3+2)2=(3)2+(2)2−2×3×2+(3)2+(2)2+2×3×2⇒3+2−26+2+3+26⇒10.\Rightarrow x^2 + \dfrac{1}{x^2} = (\sqrt{3} - \sqrt{2})^2 + (\sqrt{3} + \sqrt{2})^2 \\[1em] = (\sqrt{3})^2 + (\sqrt{2})^2 - 2 \times \sqrt{3} \times \sqrt{2} + (\sqrt{3})^2 + (\sqrt{2})^2 + 2 \times \sqrt{3} \times \sqrt{2} \\[1em] \Rightarrow 3 + 2 - 2\sqrt{6} + 2 + 3 + 2\sqrt{6} \\[1em] \Rightarrow 10.

Hence, x2+1x2=10x^2 + \dfrac{1}{x^2} = 10.

(iii) Solving,

⇒x3+1x3=(3−2)3+(3+2)3=(3)3−(2)3−3×3×2×(3−2)+(3)3+(2)3+3×3×2×(3+2)=33−22−36(3−2)+33+22+36(3+2)=33+33−22+22−318+312+318+312=63+612=63+6×23=63+123=183.\Rightarrow x^3 + \dfrac{1}{x^3} = (\sqrt{3} - \sqrt{2})^3 + (\sqrt{3} + \sqrt{2})^3 \\[1em] = (\sqrt{3})^3 - (\sqrt{2})^3 - 3 \times \sqrt{3} \times \sqrt{2} \times (\sqrt{3} - \sqrt{2}) + (\sqrt{3})^3 + (\sqrt{2})^3 + 3 \times \sqrt{3} \times \sqrt{2} \times (\sqrt{3} + \sqrt{2}) \\[1em] = 3\sqrt{3} - 2\sqrt{2} - 3\sqrt{6}(\sqrt{3} - \sqrt{2}) + 3\sqrt{3} + 2\sqrt{2} + 3\sqrt{6}(\sqrt{3} + \sqrt{2}) \\[1em] = 3\sqrt{3} + 3\sqrt{3} - 2\sqrt{2} + 2\sqrt{2} - 3\sqrt{18} + 3\sqrt{12} + 3\sqrt{18} + 3\sqrt{12} \\[1em] = 6\sqrt{3} + 6\sqrt{12} \\[1em] = 6\sqrt{3} + 6 \times 2\sqrt{3} \\[1em] = 6\sqrt{3} + 12\sqrt{3} \\[1em] = 18\sqrt{3}.

Hence, x3+1x3=183x^3 + \dfrac{1}{x^3} = 18\sqrt{3}.

(iv) Substituting values from part (i), (ii) and (iii), we get :

⇒x3+1x3−3(x2+1x2)+x+1x=183−3×10+23=203−30=10(23−3).\Rightarrow x^3 + \dfrac{1}{x^3} - 3\Big(x^2 + \dfrac{1}{x^2}\Big) + x + \dfrac{1}{x} = 18\sqrt{3} - 3 \times 10 + 2\sqrt{3} \\[1em] = 20\sqrt{3} - 30 \\[1em] = 10(2\sqrt{3} - 3).

Hence, x3+1x3−3(x2+1x2)+x+1x=10(23−3)x^3 + \dfrac{1}{x^3} - 3\Big(x^2 + \dfrac{1}{x^2}\Big) + x + \dfrac{1}{x} = 10(2\sqrt{3} - 3).

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