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SSC CGL 2023 PYQ — Question 69

SSC CGL - 26 July 2023 - Shift 3 • Quantitative Aptitude • Q69

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If (x+1x)=6\left(x + \frac{1}{x}\right) = \sqrt{6}, and x > 1, what is the value of (x8−1x8)\left(x^8 - \frac{1}{x^8}\right)?

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Correct answer: C

Explanation

The correct answer is option 3 i.e. 1123112\sqrt{3}.

We know that

(x8−1x8)\left(x^8 - \frac{1}{x^8}\right) = (x4+1x4)(x2+1x2)(x+1x)(x−1x)(x^4 + \frac{1}{x^4})(x^2+\frac{1}{x^2})(x+\frac{1}{x})(x-\frac{1}{x})

Given that

(x+1x)=6\left(x + \frac{1}{x}\right) = \sqrt{6}

Squaring both sides we get

(x2+1x2)=6−2=4(x^2+\frac{1}{x^2}) = 6-2 = 4

Again squaring both sides we get

(x4+1x4)(x^4 + \frac{1}{x^4}) = 16 - 2 = 14

Also, (x−1x)2(x-\frac{1}{x})^2 = x2+1x2−2x^2+\frac{1}{x^2} -2 = 2

⇒ (x−1x)(x-\frac{1}{x})  = √2

Hence, (x8−1x8)\left(x^8 - \frac{1}{x^8}\right) = 14 × 4 × √6 × √2 = 112√3