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SSC Selection Post 2022 PYQ — Question 71

SSC Selection Post - Matriculation Level 2022 - 14 March 2022 - Shift 1 • Quantitative Aptitude • Q71

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 If x−1x=7\sqrt x -\frac{1}{\sqrt x} =\sqrt7, then the value of x³ + 1x3\frac{1}{x^3}is:
 

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Correct answer: B

Explanation

The correct answer is Option 2 i.e. 702.

Let y = x\sqrt{x}

Then:

y−1y=7y - \frac{1}{y} = \sqrt{7}

So:

(y−1y)3=y3−1y3−3(y−1y)\left(y - \frac{1}{y}\right)^3 = y^3 - \frac{1}{y^3} - 3\left(y - \frac{1}{y}\right)

​(7)3=y3−1y3−37​(\sqrt{7})^3 = y^3 - \frac{1}{y^3} - 3\sqrt{7}

⇒ 7​7=y3−1y3−37​\sqrt{7} = y^3 - \frac{1}{y^3} - 3\sqrt{7}

⇒ 7​7+37=y3−1y3​\sqrt{7} + 3\sqrt{7} = y^3 - \frac{1}{y^3}  ⇒ 107\sqrt{7}  =y3−1y3= y^3 - \frac{1}{y^3}

⇒ y3−1y3=107y^3 - \frac{1}{y^3} = 10\sqrt{7}

⇒ (y3−1y3)2=y6+1y6−2⇒(107)2=x3+1x3−2⇒700=x3+1x3−2⇒x3+1x3=702(y^3 - \frac{1}{y^3})^2 = y^6 + \frac{1}{y^6} - 2 \Rightarrow (10\sqrt{7})^2 = x^3 + \frac{1}{x^3} - 2 \Rightarrow 700 = x^3 + \frac{1}{x^3} - 2 \Rightarrow x^3 + \frac{1}{x^3} = 702