If x = (3+8)(3 + \sqrt{8})(3+8), find the values of (x2+1x2)\Big(x^2 + \dfrac{1}{x^2}\Big)(x2+x21).
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Given,
x = (3+8)(3 + \sqrt{8})(3+8)
∴1x=1(3+8)\therefore \dfrac{1}{x} = \dfrac{1}{(3 + \sqrt{8})}∴x1=(3+8)1
Rationalizing,
⇒1x=1(3+8)×(3−8)(3−8)=3−832−(8)2=3−89−8=3−81=3−8.\Rightarrow \dfrac{1}{x} = \dfrac{1}{(3 + \sqrt{8})} \times \dfrac{(3 - \sqrt{8})}{(3 - \sqrt{8})} \\[1em] = \dfrac{3 - \sqrt{8}}{3^2 - (\sqrt{8})^2} \\[1em] = \dfrac{3 - \sqrt{8}}{9 - 8} \\[1em] = \dfrac{3 - \sqrt{8}}{1} \\[1em] = 3 - \sqrt{8}.⇒x1=(3+8)1×(3−8)(3−8)=32−(8)23−8=9−83−8=13−8=3−8.
By formula,
x2+1x2=(x+1x)2−2x^2 + \dfrac{1}{x^2} = \Big(x + \dfrac{1}{x}\Big)^2 - 2x2+x21=(x+x1)2−2
Substituting values we get :
⇒x2+1x2=(3+8+3−8)2−2=62−2=36−2=34.\Rightarrow x^2 + \dfrac{1}{x^2} = (3 + \sqrt{8} + 3 - \sqrt{8})^2 - 2 \\[1em] = 6^2 - 2 \\[1em] = 36 - 2 \\[1em] = 34.⇒x2+x21=(3+8+3−8)2−2=62−2=36−2=34.
Hence, x2+1x2x^2 + \dfrac{1}{x^2}x2+x21 = 34.
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π
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