If x+1x=x +{{1} \over x}=x+x1= 1−2,1{-\sqrt{2}},1−2, then find the value of x3+1x3.x^3+{{1} \over x^3}.x3+x31. A4−224-{2\sqrt{2}}4−22 B2−22-{\sqrt{2}}2−2 C22−4{2\sqrt{2}}-422−4 D4+224+{2\sqrt{2}}4+22 Show answer & explanationCorrect answer: AExplanationThe correct answer is Option 1 i.e. 4−224-{2\sqrt{2}}4−22 x+1x=1−2x + \frac{1}{x} = 1 - \sqrt{2} x+x1=1−2 x+1x=ax + \frac{1}{x} = a x+x1=a a=1−2a = 1 - \sqrt{2} \\a=1−2 x3+1x3=a3−3ax^3 + \frac{1}{x^3} = a^3 - 3ax3+x31=a3−3a a3=(1−2)3=13−3(12)(2)+3(1)(2)2−(2)3=1−32+6−22=7−52a^3= (1 - \sqrt{2})^3 = 1^3 - 3(1^2)(\sqrt{2}) + 3(1)(\sqrt{2})^2 - (\sqrt{2})^3 = 1 - 3\sqrt{2} + 6 - 2\sqrt{2} = 7 - 5\sqrt{2}a3=(1−2)3=13−3(12)(2)+3(1)(2)2−(2)3=1−32+6−22=7−52 x3+1x3=(7−52)−3(1−2)=7−52−3+32=4−22x^3 + \frac{1}{x^3} = (7 - 5\sqrt{2}) - 3(1 - \sqrt{2}) = 7 - 5\sqrt{2} - 3 + 3\sqrt{2} = 4 - 2\sqrt{2}x3+x31=(7−52)−3(1−2)=7−52−3+32=4−22