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SSC CGL 2023 PYQ — Question 59

SSC CGL - 21 July 2023 - Shift 2 • Quantitative Aptitude • Q59

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If (x+1x)=22(x+\frac{1}{x})=2\sqrt{2},  what is the value of (x6+1x6)?(x^6+\frac{1}{x^6})?

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Correct answer: D

Explanation

The correct answer is option 4 i.e. 198.

(x+1x)=22(x+\frac{1}{x})=2\sqrt{2}

⇒ (x2+1x2)(x^2+\frac{1}{x^2}) = (22)2−2(2\sqrt{2})^2-2 = 6

Now, cube on both sides

⇒ x6+1x6x^6+\frac{1}{x^6} + 3(x2+1x2)(x^2+\frac{1}{x^2}) = 216

⇒ x6+1x6x^6+\frac{1}{x^6} + 18 = 216

⇒ x6+1x6x^6+\frac{1}{x^6} = 216 - 18 = 198