If x-\frac{1}{x}=\sqrt{77}, then one of the values of x^{3}+\frac{1}{x^{3}} is

SSC CGL 18 Aug 2021 Shift 3 Paper

Question : 64 of 100

Marks: +1, -0

x-\;{1}/{x}=√{77}

, then one of the values of

x^{3}+\;{1}/{x^{3}}

Solution:

Given,

x-\;{1}/{x}=√{77}

Take square on both sides,

(x-\;{1}/{x})^{2}=(√{77})^{2}

⇒ \;\; x^{2}+\;{1}/{x^{2}}-2=(√{77})^{2}

⇒ \;\; x^{2}+\;{1}/{x^{2}}+2-2-2=(√{77})^{2}

⇒ \;\; (√{77})^{2}=(x+\;{1}/{x})^{2}-4

⇒ \;\; (x+\;{1}/{x})=√{81}=9

⇒ \;\; (x+\;{1}/{x})^{2}

⇒ \;\; (x+\;{1}/{x}=(x)

Take Cube on both sides,

x^{3}+\;{1}/{x^{3}}+3 x × \;{1}/{x}(x+\;{1}/{x})=9 × 9 × 9

{[ ∵ (a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)]}

⇒ x^{3}+\;{1}/{x^{3}}=729-3 × 9=729-27

∴ x^{3}+\;{1}/{x^{3}}=702 .

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