If x^{4}-62 x^{2}+1=0, where x>0, then the value of x^{3}+x^{-3} is

SSC CGL 24 Aug 2021 Shift 3 Paper

Question : 52 of 100

Marks: +1, -0

x^{4}-62 x^{2}+1=0

, where

x>0

, then the value of

x^{3}+x^{-3}

Solution:

Given,

x^{4}-62 x^{2}+1=0

x^{4}+1=62 x^{2}

Divided by

x^{2}

, we get

x^{2}+\;{1}/{x^{2}}=62

We know that,

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

(x+\;{1}/{x})^{2}=62+2=64

{[ ∵ (a+b)^{2}=a^{2}+b^{2}+2 a b]}

(x+\;{1}/{x})=√{64}=8 ⇒(x+\;{1}/{x})=8

Taking cube on both sides,

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

× \;{1}/{x}(x+\;{1}/{x})=8^{3}

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

⇒ x^{3}+\;{1}/{x^{3}}=512-3 × 8=512-24

⇒ x^{3}+\;{1}/{x^{3}}=488

x^{3}+x^{-3}=488

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