If (x+\frac{1}{x})^{3}=27, then what is the value of (x^{2}+\frac{1}{x^{2}}) ? Given that x is real.

SSC CHSL 4 Aug 2021 Shift 1 Paper

Section: Quantitative Aptitude

Question : 69 of 100

Marks: +1, -0

(x+\;{1}/{x})^{3}=27

, then what is the value of

(x^{2}+\;{1}/{x^{2}})

? Given that

x

is real.

Solution:

Given,

(x+\;{1}/{x})^{3}=27

⇒ \;\; x+\;{1}/{x}=(27)^{\;{1}/{3}}=3

∵ \;\; (a+b)^{2}=a^{2}+b^{2}+2 a b

∴ (x+\;{1}/{x})^{2}=x^{2}+\;{1}/{x^{2}}+2 × x × \;{1}/{x}

⇒ \;\; (3)^{2}=x^{2}+\;{1}/{x^{2}}+2

∴ x^{2}+\;{1}/{m^{2}}=9-2=7

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