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SSC CGL 18 Aug 2021 Shift 3 Paper

Question : 64 of 100

Marks: +1, -0

x - \frac{1}{x} = \sqrt{77}

x^{3} + \frac{1}{x^{3}}

Solution: 👈: Video Solution

Given,

x - \frac{1}{x} = \sqrt{77}

Take square on both sides,

\left(x - \frac{1}{x}\right)^{2} = (\sqrt{77})^{2}

\Rightarrow \;\; x^{2} + \frac{1}{x^{2}} - 2 = (\sqrt{77})^{2}

\Rightarrow \;\; x^{2} + \frac{1}{x^{2}} + 2 - 2 - 2 = (\sqrt{77})^{2}

\Rightarrow \;\; (\sqrt{77})^{2} = \left(x + \frac{1}{x}\right)^{2} - 4

\Rightarrow \;\; \left(x + \frac{1}{x}\right) = \sqrt{81} = 9

\Rightarrow \;\; \left(x + \frac{1}{x}\right)^{2}

\Rightarrow \;\; (x + \frac{1}{x} = (x)

Take Cube on both sides,

x^{3} + \frac{1}{x^{3}} + 3 x \times \frac{1}{x} \left(x + \frac{1}{x}\right) = 9 \times 9 \times 9

\left[ \because (a+b)^{3} = a^{3} + b^{3} + 3ab(a+b) \right]

\Rightarrow x^{3} + \frac{1}{x^{3}} = 729 - 3 \times 9 = 729 - 27

\therefore x^{3} + \frac{1}{x^{3}} = 702.

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